Differentiation (Part-38) | S N De

Tangent and Normal | Part-7

 

In the previous article , we have solved few Short Answer Type Questions. In the following article, we are going to discuss/solve few more  Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Tangent and Normal (Ex-14).

14. If the tangent to the curve $~y=x^3+ax+b~$ at $~(1,-6)~$ is parallel to the line $~x-y+5=0,~$ find $~a~$ and $~b.$ 

Solution.

$y=x^3+ax+b \longrightarrow(1) \\ \therefore~\frac{dy}{dx}=3x^2+a \\ \text{So,}~~ \left[\frac{dy}{dx}\right]_{(1,-6)}=3 \cdot 1^2+a=3+a$

Slope of the given straight line $~x-y+5=0~$ is $~1.$

Clearly, by question

$3+a=1 \Rightarrow a=-2.$

Since the point $~(1,-6)~$ lies on the curve (1), so

$~-6=1^3+a \cdot 1+b \\ \text{or,}~~ -6=1+(-2)+b \Rightarrow b=-5.$

$\therefore~~ a=-2,~~ b=-5~~\text{(ans)}$

15. The slope of the tangent to the parabola $~3y^2=8x~$ at the point $~\left(\frac 23 t^2,\frac 43t\right)~$ is $~(-2);~$ find the equation of the tangent.

Solution.

$3y^2=8x \longrightarrow(1)$

Differentiating w.r.t. $~x~$ , we get

$3 \times 2y~\frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx}=\frac{8}{3 \times 2y}=\frac{4}{3y}.$

By question,

$\left[\frac{dy}{dx}\right]_{\left(\frac 23 t^2,\frac 43 t\right)}=-2 \\ \text{or,}~~ \frac{4}{3 \times \frac 43 t}=-2 \\ \text{or,}~~ \frac{4}{4t}=-2 \\ \text{or,}~~ t=-\frac 12$

So, coordinates of point of contact of the tangent 

$\left(\frac 23 \cdot (-1/2)^2, \frac 43 \cdot (-1/2)\right)=\left(\frac 16, -\frac 23\right).$

Hence, the equation of tangent to the given parabola (1) at $~\left(\frac 16, -\frac 23\right)~$ is 

$y-\left(-\frac 23\right)=-2\left(x-\frac 16\right) \\ \text{or,}~~ y+\frac 23=-2x+\frac 13 \\ \text{or,}~~ 3y+2=-6x+1 \\ \therefore~~ 6x+3y+1=0~~\text{(ans)}$

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16. Find the equations of the tangents to the circle $~x^2+y^2 = 16~$

having slope $~\left(-\frac 43\right)$. 

Solution.

$x^2+y^2=16 \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$ , we get

$2x+2y~\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{2x}{2y}=-\frac xy \rightarrow(2)$

Let the slope of the tangent to the circle (1) at $~(h,k)~$ be $~-\frac 43.$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(h,k)}=-\frac 43 \\ \text{or,}~~ -\frac hk=-\frac 43~~[\text{By (2)}] \Rightarrow k=\frac{3h}{4}\rightarrow(3)$

Now, since the point $~(h,k)~$ lies on the circle (1), so using (3) we get

$h^2+k^2=16 \\ \text{or,}~~ h^2+\left(\frac{3h}{4}\right)^2=16 \\ \text{or,}~~ h^2+\frac{9h^2}{16}=16 \\ \text{or,}~~ \frac{16h^2+9h^2}{16}=16\\ \text{or,}~~ \frac{25h^2}{16}=16 \\ \text{or,}~~ h=\pm \sqrt{\frac{16 \times 16}{25}}=\pm \frac{16}{5}$

$\therefore~~ k=\frac 34 \times \left( \pm \frac{16}{5}\right)=\pm \frac{12}{5}.$

So, the equation of tangent to the given circle at $~\left(\frac{16}{5},\frac{12}{5}\right)~$ is

$y-\frac{12}{5}=-\frac 43\left(x-\frac{12}{5}\right) \\ \text{or,}~~ \frac{5y-12}{5}=-\frac{4x}{3}+\frac{64}{15} \\ \text{or,}~~ \frac{5y-12}{5}=\frac{-20x+64}{15} \\ \text{or,}~~ 5y-12=\frac 13(-20x+64) \\ \text{or,}~~ 15y-36=-20x+64 \\ \text{or,}~~ 20x+15y=64+36 \\ \text{or,}~~ 5(4x+3y)=100 \\ \text{or,}~~ 4x+3y=20$

Again, the equation of the curve to the given circle (1) at $~\left(-\frac{16}{5},-\frac{12}{5}\right)~$ is

$y-\left(-\frac{12}{5}\right)=-\frac 43 \left[x-\left(-\frac{16}{5}\right)\right] \\ \text{or,}~~ y+\frac{12}{5}=-\frac{4x}{3}-\frac{64}{15} \\ \text{or,}~~ \frac{5y+12}{5}=\frac{-20x-64}{15} \\ \text{or,}~~ 5y+12=\frac{-20x-64}{3} \\ \text{or,}~~ 15y+36=-20x-64 \\ \text{or,}~~ 20x+15y=-36-64 \\ \text{or,}~~ 4(4x+3y)=-100 \\ \text{or,}~~ 4x+3y=-20$

Hence, required equations of tangents having slope $~-\frac 43~$ are 

$4x+3y=\pm 20$

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17. The slope of tangent to the ellipse $~x^2+4y^2 = 4~$ at the point

$~(2 \cos\theta, \sin\theta)~$ is $~\sqrt{2}~$; find the equation of the tangent. 

Solution.

$x^2+4y^2=4\longrightarrow(1) $

Differentiating (1) w.r.t. $~x~$ , we get

$2x+8y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{2x}{8y}=-\frac{x}{4y}$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(2\cos\theta,\sin\theta)}=-\frac{2\cos\theta}{4\sin\theta}=-\frac 12 \cot\theta$

So, the equation of tangent to the given curve (1) at $~(2\cos\theta,\sin\theta)~$ is 

$y-\sin\theta=-\frac 12 \cot\theta (x-2\cos\theta) \\ \text{or,}~~ y=-\frac 12x \cot\theta+ \cos\theta \cot \theta+\sin\theta \\ \text{or,}~~ y=-\frac 12x+\cos\theta \cdot \frac{\cos\theta}{\sin\theta}+\sin\theta \\ \text{or,}~~ y=-\frac 12x \cot\theta+\frac{\cos^2\theta}{\sin\theta}+\sin\theta \\ \text{or,}~~ y=-\frac 12x\cot\theta+\frac{\cos^2\theta+\sin^2\theta}{\sin\theta} \\ \text{or,}~~ y=-\frac 12x\cot\theta+\frac{1}{\sin\theta} \\ \text{or,}~~ y=-\frac 12x\cot\theta+\csc\theta$

By question,

$-\frac 12 \cot\theta=\sqrt{2} \\ \text{or,}~~ \cot\theta=\sqrt{2} \times (-2)=-2\sqrt{2}$

$\csc\theta=\pm \sqrt{1+\cot^2\theta}=\pm\sqrt{1+8}=\pm 3$

Hence, the equations of tangents are $~y=\sqrt{2} x \pm 3.$ 

18. Find the equation of the tangent to the curve $~y = \sqrt{3x-2}~$

which is parallel to the line $~4x-2y+5=0$. [NCERT, CBSE '09]

Solution.

$y=\sqrt{3x-2} \longrightarrow(1) \\ \text{or,}~~ y^2=3x-2 \\ \therefore~~ 2y~\frac{dy}{dx}=3 \Rightarrow \frac{dy}{dx}=\frac{3}{2y}.$

Let the tangent to the given curve at $~(h,k)~$ be parallel to the line $~4x-2y+5=0 \rightarrow(2)$

From (2) we get, $~y=2x+\frac 52~$ and so the slope of the straight line (2) is $~2.$

By question, we get

$~\left[\frac{dy}{dx}\right]_{(h,k)}=2 \\ \text{or,}~~ \frac{3}{2k}=2 \Rightarrow k=\frac 34.$

Since the point $~(h,k)~$ lies on the curve (1), so

$k=\sqrt{3h-2} \\ \text{or,}~~ k^2=3h-2 \\ \text{or,}~~ 3h=\left(\frac 34\right)^2 +2 =\frac{9}{16}+2 \\ \text{or,}~~ 3h=\frac{9+32}{16} \\ \therefore~~ h=\frac{41}{48}.$

So, the equation of tangent to the given curve at $~\left(\frac{41}{48},\frac 34\right)~$ is 

$y-\frac 34=2 \left(x-\frac{41}{48}\right) \\ \text{or,}~~ 4y-3=2 \times 4\left(x-\frac{41}{48}\right)\\ \text{or,}~~ 4y-3=8x-\frac{41}{6} \\ \text{or,}~~ 24y-18=48x-41 \\ \therefore~~ 48x-24y=23~~\text{(ans)} $

19. Find the points on the curve $~y=x^3~$ where the slope of the tangent is equal to x-coordinate of the point. [CBSE '08] 

Solution.

$y=x^3 \longrightarrow(1) \\ \therefore~~ \frac{dy}{dx}=3x^2$

Let the coordinates of the required point $~(h,k).$

$\therefore~~$ By question, $~\left[\frac{dy}{dx}\right]_{(h,k)}=h$

$ \text{So,}~~ 3h^2=h \\ \text{or,}~~ h(3h-1)=0 \\ \text{or,}~~ h=0, 3h-1=0 \\ \therefore~~ h=0,\frac 13.$

Since the point $~(h,k)~$ lies on the curve (1), $~k=h^3.$

For $~h=0,~~k=0.$

For $~h=\frac 13,~~ k=\left(\frac 13\right)^3=\frac{1}{27}.$

Hence, the coordinates of the required points are $~(0,0)~$ and $~\left(\frac{1}{3},\frac{1}{27}\right).$

20. Using calculus find the coordinates of the point on the parabola $~y^2 = 12x~$ at which the tangent is parallel to the line $~2x+3y=5.$

Solution.

$y^2=12x \longrightarrow(1) \\ \therefore~~ 2y~\frac{dy}{dx}=12 \\ \text{or,}~~ \frac{dy}{dx}=\frac{12}{2y}=\frac 6y.$

Let the tangent to the given curve (1) at $~(h,k)~$ be parallel to the line $~2x+3y=5 \rightarrow (2)$

From (2) we get, $~y=-\frac 23x+\frac 53.$

Now, the slope of the straight line (2) is $~ -\frac 23.$

By question, $~ \left[\frac{dy}{dx}\right]_{(h,k)}=-\frac 23$

$\text{So,}~~ \frac 6k=-\frac 23 \Rightarrow k=-\frac{6 \times 3}{2}=-9.$

Since the point $~(h,k)~$ lies on the curve (1), $~k^2=12h \Rightarrow h=\frac{k^2}{12}.$

For $~ k=-9,~~ h=\frac{(-9)^2}{12}=\frac{81}{12}=\frac{27}{4}.$

Hence, the coordinates of the required point is $~\left(\frac{27}{4},-9\right).$

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21. Find the equation of the tangent to the ellipse $~4x^2 +9y^2 = 36~$ at the point $~(x_1,y_1)~$; hence, find the coordinates of the points on this ellipse at which the tangents are parallel to the line $~2x-3y=6$.

Solution.

$4x^2+9y^2=36 \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$, we get

$8x+18y~\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{8x}{18y}=-\frac{4x}{9y}.$

The equation of the tangent to the given curve at $~(x_1,y_1)~$ is given by 

$y-y_1=\left[\frac{dy}{dx}\right]_{(x_1,y_1)}(x-x_1) \\ \text{or,}~~ y-y_1=-\frac{4x_1}{9y_1}(x-x_1) \\ \text{or,}~~ 9yy_1-9y_1^2=-4xx_1+4x_1^2 \\ \text{or,}~~ 4xx_1+9yy_1=4x_1^2+9y_1^2 \\ \therefore~~ 4xx_1+9yy_1=36~~[*]$

Note[*] : Since the point $~(x_1,y_1)~$ lies on (1),  $~4x_1^2+9y_1^2=26.$

2nd Part :

Let us assume that the tangent to the given curve at $~(h,k)~$ is parallel to the straight line $~2x-3y=6 \rightarrow(2)$

From (2) we get, $~y=\frac 23x-2.$

So, the slope of the straight line (2) is $~\frac 23.$

By question, $~\left[\frac{dy}{dx}\right]_{(h,k)}=\frac 23$

$\text{So,}~~ -\frac{4h}{9k}=\frac 23 \Rightarrow h=-\frac 32k.$

Since the point $~(h,k)~$ lies on the ellipse (1), so

$4h^2+9k^2=36 \\ \text{or,}~~ 4 \times \left(-\frac 32k\right)^2+9k^2=36 \\ \text{or,}~~ 4 \times \frac 94k^2+9k^2=36 \\ \text{or,}~~ 18k^2=36 \\ \therefore~~k=\pm \sqrt{\frac{36}{18}}=\pm \sqrt{2}.$

For $~k=\sqrt{2},~~ h=-\frac 32 \cdot \sqrt{2}=-\frac{3\sqrt{2}}{2}$

For $~k=-\sqrt{2},~~ h=-\frac 32 \cdot (-\sqrt{2})=\frac{3\sqrt{2}}{2}$

Hence, the coordinates of the required points are $~\left(-\frac{3\sqrt{2}}{2},\sqrt{2}\right)$ and $~\left(\frac{3\sqrt{2}}{2},-\sqrt{2}\right)$.

22. Find the equation of the tangent to the parabola $y^2 = 8x,~$ which is inclined at an angle $~45^{\circ}~$ with the x-axis.

Solution.

$y^2=8x \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$ , we get

$2y~\frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx}=\frac{8}{2y}=\frac 4y.$

Let us assume that the tangent to the given curve at $~(h,k)~$ is inclined at an angle $~45^{\circ}~$ with the $x-$axis.

$\therefore~~ \left[\frac{dy}{dx}\right]_{(h,k)}=\tan 45^{\circ} \\ \text{or,}~~ \frac 4k=1 \Rightarrow k=4.$

Since the point $~(h,k)~$ lies on the given parabola at $~(2,4)~$ is 

$y-4=1 \cdot (x-2) \Rightarrow y=x+2.$

23. Find the equation of the tangent to the ellipse $~x^2 + 16y^2 = 16~$ at the point $~(4 \cos\alpha, \sin\alpha)~$. Hence, find the equations of the tangents to the ellipse which are inclined at an angle $~60^{\circ}$ to x-axis.

Solution.

$x^2+16y^2=16 \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$ , we get

$2x+32y~\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac{2x}{32y}=-\frac{x}{16y}$

$\left[\frac{dy}{dx}\right]_{(4\cos\alpha,\sin\alpha)}=-\frac{4\cos\alpha}{16\sin\alpha}=-\frac{\cos\alpha}{4\sin\alpha}$

So, the equation of tangent at $~(4\cos\alpha,\sin\alpha)~$ is 

$y-\sin\alpha=-\frac{\cos\alpha}{4\sin\alpha}(x-4\cos\alpha) \\ \text{or,}~~ 4y\sin\alpha-4\sin^2\alpha=-x\cos\alpha+4\cos^2\alpha \\ \text{or,}~~ x\cos\alpha+4y\sin\alpha=4(\sin^2\alpha+\cos^2\alpha)=4\rightarrow (2)$

If the tangent is inclined at an angle $~60^{\circ}~$ with the x-axis, then 

$-\frac{\cos\alpha}{4\sin\alpha}=\tan 60^{\circ}=\sqrt{3} \\ \text{or,}~~ \cot\alpha=-4\sqrt{3} \\ \therefore~~ \csc\alpha=\mp\sqrt{1+\cot^2\alpha}=\mp \sqrt{1+(-4\sqrt{3})^2} \\ \text{or,}~~ \csc\alpha=\mp \sqrt{1+48}=\mp 7.$

From (2) we get,

$x\cot\alpha+4y=4\csc\alpha \\ \text{or,}~~ x(-4\sqrt{3})+4y=4 \times (\mp 7) \\ \text{or,}~~ -4(\sqrt{3} x-y)= \mp 4 \times 7 \\ \therefore~~ \sqrt{3} x-y= \pm 7~~\text{(ans)}$

24. Find the equations of the tangents to the hyperbola $~3x^{2}-4y^2= 12,~$ which are inclined at an angle $~60^{\circ}$ to x-axis.

Solution.

$3x^2-4y^2=12 \rightarrow(1)$

Differentiating (1) w.r.t. $x$, we get

$6x-8y~\frac{dy}{dx}=0\\~~ \Rightarrow \frac{dy}{dx}=\frac{6x}{8y}=\frac{3x}{4y}$

Let us consider the tangent to (1) at $~(h,k)~$ makes an angle $~60^{\circ}~$ with $x-$ axis.

$\therefore~~\left[\frac{dy}{dx}\right]_{(h,k)}=\tan 60^{\circ}\\~~\Rightarrow \frac{3h}{4k}= \sqrt{3}$

$\therefore~3h=4k\sqrt{3} \Rightarrow h=\frac{4k\sqrt{3}}{3}=\frac{4k}{\sqrt{3}} \rightarrow(2)$

Since the point $~(h,k)~$ lies on the given curve (1), so

$3h^2-4k^2=12 \\ \text{or,}~~ 3 \times \left(\frac{4k}{\sqrt{3}}\right)^2-4k^2=12 \\ \text{or,}~~ 16k^2-4k^2=12 \\ \text{or,}~~ 12k^2=12 \\ \text{or,}~~ k^2=1\Rightarrow k=\pm 1.$

$\therefore~~h= \frac{4}{\sqrt{3}} \times (\pm 1)=\pm \frac{4}{\sqrt{3}}~~[\text{By (2)}]$

Hence, the equation of the tangent at the point $~\left( \pm \frac{4}{\sqrt{3}}, \pm 1\right)~$ is given by 

$y- (\pm 1)=\sqrt{3} \left[x-\left( \pm \frac{4}{\sqrt{3}}\right)\right] \\ \text{or,}~~ y \mp 1=\sqrt{3}x \mp 4 \\ \text{or,}~~ y-\sqrt{3}x=\mp 4-(\mp 1) \\ \therefore~~y=\sqrt{3}x \pm 3.$

25. Prove that the lengths of the tangents from any point on the line $~3x-8y+2=0~$ to the circles $~x^2+y^2+2x-10y+12=0~$ and $~x^2+y^2-4x+6y+8 = 0~$ are equal.

Solution.

Let $(\alpha,\beta)$ be any point on the straight line $~3x-8y+2=0.$

$\therefore~~ 3\alpha-8\beta+2=0 \Rightarrow \beta=\frac{3\alpha+2}{8} \rightarrow(1)$

The given circles are 

$x^2+y^2+2x-10y+12=0 \rightarrow(2),$

$x^2+y^2-4x+6y+8=0 \rightarrow(3).$

Now, the length of the tangent from the point $~(\alpha,\beta)~$ to the circle (2) is given by 

$~\sqrt{\alpha^2+\beta^2+2\alpha-10\beta+12}=T \\ \text{or,}~~ T^2 \\=\alpha^2+\left(\frac{3\alpha+2}{8}\right)^2+2\alpha-10 \times \frac{3\alpha+2}{8}+12 \\=\alpha^2+\frac{1}{64}(9\alpha^2+12\alpha+4)\\~~+2\alpha-\frac 54(3\alpha+2)+12 \\=\frac{1}{64}(64\alpha^2+9\alpha^2+12\alpha+4\\~~+128\alpha-240\alpha-160+768)\\=\frac{1}{64}(73\alpha^2-100\alpha+612) \rightarrow(4)$

Again, the length of the tangent from the point $~(\alpha,\beta)~$ to the circle (3) is given by 

$S^2=\alpha^2+\beta^2-4\alpha +6\beta+8\\=\alpha^2+\left(\frac{3\alpha+2}{8}\right)^2-4\alpha+6 \times \frac{3\alpha+2}{8}+8 \\= \alpha^2+\frac{1}{64}(9\alpha^2+12\alpha+4)-4\alpha\\~~+\frac 18(18\alpha+12)+8\\=\frac{1}{64}(64\alpha^2+9\alpha^2+12\alpha+4\\~~-256\alpha+144\alpha+96+512)\\=\frac{1}{64}(73\alpha^2-100\alpha+612)\rightarrow(5)$

From (4) and (5), we can conclude that 

$T^2=S^2 \Rightarrow T=S~( \because ~ T,S >0)$

Hence, the length of the tangents are equal.

26. If the length of the tangent drawn from $~(f, g)~$ to the circle $~x^2+ y^2 = 6~$ be twice the length of the tangent drawn from the same point to the circle $~x^2 + y^2+3(x+y)= 0~$ then show that $~g^2+f^2+4g+4f+2=0$.

Solution.

The length of the tangent from the point $~(f,g)~$ to the circle $~x^2+y^2-6=0~$ is $~\sqrt{f^2+g^2-6}~$ unit.

Again, the length of the tangent from the point $~(f,g)~$ to the circle $~x^2+y^2+3(x+y)=0~$ is given by $~\sqrt{f^2+g^2+3f+3g}~$ unit.

So, by question,

$~\sqrt{f^2+g^2-6}=2\sqrt{f^2+g^2+3f+3g} \\ \text{or,}~~ f^2+g^2-6=4(f^2+g^2+3f+3g) \\ \text{or,}~~ f^2-4f^2+g^2-4g^2-12f-12g-6=0 \\ \text{or,}~~ -3f^2-3g^2-12f-12g-6=0 \\ \text{or,}~~ -3(f^2+g^2+4f+4g+2)=0 \\ \therefore~~ f^2+g^2+4f+4g+2=0.$