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:
Given curve equation: $y=x^3+ax+b \quad \longrightarrow(1)$
Differentiating w.r.t. $x$: $\frac{dy}{dx}=3x^2+a$
Slope of tangent at $(1,-6)$: $\left[\frac{dy}{dx}\right]_{(1,-6)} = 3(1)^2+a = 3+a$
Slope of given line $x-y+5=0$ is $1$.
Since the tangent is parallel to the line: $3+a=1 \implies a=-2$
Since $(1,-6)$ lies on curve (1): $-6 = (1)^3 + a(1) + b$
$-6 = 1 + (-2) + b \implies b=-5$
$\therefore a = -2, \quad b = -5$
15. The slope of the tangent to the parabola $3y^2=8x$ at the point $\left(\frac{2}{3}t^2,\frac{4}{3}t\right)$ is $(-2)$; find the equation of the tangent.
Solution:
$3y^2=8x \quad \longrightarrow(1)$
Differentiating w.r.t. $x$: $3 \times 2y \frac{dy}{dx}=8 \implies \frac{dy}{dx}=\frac{4}{3y}$
According to the problem: $\left[\frac{dy}{dx}\right]_{\left(\frac{2}{3}t^2,\frac{4}{3}t\right)} = -2 \implies \frac{4}{3 \times \frac{4}{3}t} = -2 \implies t = -\frac{1}{2}$
Coordinates of point of contact: $\left(\frac{2}{3}\left(-\frac{1}{2}\right)^2, \frac{4}{3}\left(-\frac{1}{2}\right)\right) = \left(\frac{1}{6}, -\frac{2}{3}\right)$
Equation of tangent at $\left(\frac{1}{6}, -\frac{2}{3}\right)$:
$y-\left(-\frac{2}{3}\right) = -2\left(x-\frac{1}{6}\right) \implies y+\frac{2}{3} = -2x+\frac{1}{3}$
$\therefore 6x+3y+1=0$
16. Find the equations of the tangents to the circle $x^2+y^2 = 16$ having slope $\left(-\frac{4}{3}\right)$.
Solution:
$x^2+y^2=16 \quad \longrightarrow(1)$
Differentiating (1) w.r.t. $x$: $2x+2y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{x}{y} \quad \longrightarrow(2)$
Let tangent at $(h,k)$ have slope $-\frac{4}{3}$.
$\left[\frac{dy}{dx}\right]_{(h,k)} = -\frac{h}{k} = -\frac{4}{3} \implies k = \frac{3h}{4} \quad \longrightarrow(3)$
Since $(h,k)$ lies on (1): $h^2+\left(\frac{3h}{4}\right)^2 = 16 \implies \frac{25h^2}{16}=16 \implies h=\pm\frac{16}{5}$
$\therefore k = \pm\frac{12}{5}$
At $\left(\frac{16}{5}, \frac{12}{5}\right)$, equation is: $y-\frac{12}{5} = -\frac{4}{3}\left(x-\frac{16}{5}\right) \implies 4x+3y=20$
At $\left(-\frac{16}{5}, -\frac{12}{5}\right)$, equation is: $y+\frac{12}{5} = -\frac{4}{3}\left(x+\frac{16}{5}\right) \implies 4x+3y=-20$
$\therefore 4x+3y = \pm 20$
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:
$x^2+4y^2=4 \quad \longrightarrow(1)$
Differentiating w.r.t $x$: $2x+8y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{x}{4y}$
$\left[\frac{dy}{dx}\right]_{(2\cos\theta, \sin\theta)} = -\frac{2\cos\theta}{4\sin\theta} = -\frac{1}{2}\cot\theta$
Equation of tangent at $(2\cos\theta, \sin\theta)$: $y = -\frac{1}{2}x\cot\theta + \csc\theta$
Given slope: $-\frac{1}{2}\cot\theta = \sqrt{2} \implies \cot\theta = -2\sqrt{2}$
$\csc\theta = \pm\sqrt{1+\cot^2\theta} = \pm\sqrt{1+8} = \pm 3$
$\therefore y = \sqrt{2}x \pm 3$
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:
$y^2=3x-2 \implies 2y\frac{dy}{dx}=3 \implies \frac{dy}{dx}=\frac{3}{2y}$
Slope of line $4x-2y+5=0 \implies y=2x+\frac{5}{2}$ is $2$.
At point $(h,k)$: $\frac{3}{2k} = 2 \implies k = \frac{3}{4}$
Since $(h,k)$ lies on curve: $3h = k^2+2 = \left(\frac{3}{4}\right)^2+2 = \frac{41}{16} \implies h = \frac{41}{48}$
Equation of tangent at $\left(\frac{41}{48}, \frac{3}{4}\right)$: $y-\frac{3}{4} = 2\left(x-\frac{41}{48}\right)$
$\therefore 48x - 24y = 23$
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:
$y=x^3 \implies \frac{dy}{dx}=3x^2$
Let required point be $(h,k)$. Given: $\left[\frac{dy}{dx}\right]_{(h,k)} = h \implies 3h^2 = h$
$h(3h-1)=0 \implies h=0 \text{ or } h=\frac{1}{3}$
For $h=0 \implies k=0^3=0$
For $h=\frac{1}{3} \implies k=\left(\frac{1}{3}\right)^3 = \frac{1}{27}$
Required points: $(0,0)$ 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:
$y^2=12x \implies 2y\frac{dy}{dx}=12 \implies \frac{dy}{dx}=\frac{6}{y}$
Slope of line $2x+3y=5 \implies y=-\frac{2}{3}x+\frac{5}{3}$ is $-\frac{2}{3}$.
At point $(h,k)$: $\frac{6}{k} = -\frac{2}{3} \implies k = -9$
Since $(h,k)$ lies on parabola: $h = \frac{k^2}{12} = \frac{(-9)^2}{12} = \frac{27}{4}$
Required point: $\left(\frac{27}{4}, -9\right)$
21. Find the equation of the tangent to the ellipse $4x^2 +9y^2 = 36$ at $(x_1,y_1)$; hence, find the coordinates of the points on this ellipse at which the tangents are parallel to the line $2x-3y=6$.
Solution:
Part 1: $4x^2+9y^2=36 \implies \frac{dy}{dx} = -\frac{4x}{9y}$
Tangent at $(x_1,y_1)$: $y-y_1 = -\frac{4x_1}{9y_1}(x-x_1) \implies 4xx_1 + 9yy_1 = 4x_1^2+9y_1^2$
Since $(x_1,y_1)$ lies on ellipse, $4x_1^2+9y_1^2=36 \implies \mathbf{4xx_1+9yy_1=36}$
Part 2: Parallel to line $2x-3y=6$ (Slope $= \frac{2}{3}$).
At point $(h,k)$: $-\frac{4h}{9k} = \frac{2}{3} \implies h = -\frac{3}{2}k$
Since $(h,k)$ lies on ellipse: $4\left(-\frac{3}{2}k\right)^2 + 9k^2 = 36 \implies 18k^2 = 36 \implies k = \pm\sqrt{2}$
If $k = \sqrt{2} \implies h = -\frac{3\sqrt{2}}{2}$; If $k = -\sqrt{2} \implies h = \frac{3\sqrt{2}}{2}$
Required points: $\left(-\frac{3\sqrt{2}}{2}, \sqrt{2}\right)$ 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:
$y^2=8x \implies 2y\frac{dy}{dx}=8 \implies \frac{dy}{dx}=\frac{4}{y}$
Given angle $\theta = 45^\circ \implies \text{Slope } m = \tan 45^\circ = 1$
At point $(h,k)$: $\frac{4}{k} = 1 \implies k=4$
Since $(h,k)$ lies on parabola: $h = \frac{k^2}{8} = \frac{16}{8} = 2$
Equation of tangent at $(2,4)$: $y-4 = 1(x-2)$
$\therefore y = x + 2$
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:
Part 1: $x^2+16y^2=16 \implies \frac{dy}{dx} = -\frac{x}{16y}$
Slope at $(4\cos\alpha, \sin\alpha)$: $\left[\frac{dy}{dx}\right] = -\frac{4\cos\alpha}{16\sin\alpha} = -\frac{1}{4}\cot\alpha$
Equation of tangent: $y-\sin\alpha = -\frac{\cos\alpha}{4\sin\alpha}(x-4\cos\alpha) \implies \mathbf{x\cos\alpha + 4y\sin\alpha = 4}$
Part 2: Angle with x-axis is $60^\circ \implies \text{Slope } = \tan 60^\circ = \sqrt{3}$
$-\frac{1}{4}\cot\alpha = \sqrt{3} \implies \cot\alpha = -4\sqrt{3}$
$\csc\alpha = \pm\sqrt{1+\cot^2\alpha} = \pm\sqrt{1+48} = \pm 7 \implies \sin\alpha = \pm\frac{1}{7}$
$\cos\alpha = \cot\alpha \cdot \sin\alpha = \mp\frac{4\sqrt{3}}{7}$
Substituting $\cos\alpha$ & $\sin\alpha$ into the tangent equation gives:
$\therefore y = \sqrt{3}x \pm 7$
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