Differentiation (Part-38) | S N De

Tangent and Normal VSA type Solutions

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

Find the equation of the tangent at the specified points to each of the following curves:

5. The parabola $y^2=4x$ at $(1,2)$.

Solution:

The equation of the curve is:
$$ y^2 = 4x \quad \dots(1) $$

Differentiating both sides of (1) with respect to $x$, we get:
$$ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} $$

At point $(1,2)$:
$$ \left[\frac{dy}{dx}\right]_{(1,2)} = \frac{2}{2} = 1 $$

The equation of the tangent at $(1,2)$ is given by:
$$ y - 2 = \left[\frac{dy}{dx}\right]_{(1,2)} (x - 1) $$
$$ y - 2 = 1 \cdot (x - 1) $$
$$ y - 2 = x - 1 \implies x - y + 1 = 0 $$

Answer: $x - y + 1 = 0$

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6. The ellipse $9x^2+16y^2=288$ at $(4,3)$.

Solution:

The equation of the curve is:
$$ 9x^2 + 16y^2 = 288 \quad \dots(1) $$

Differentiating both sides of (1) w.r.t. $x$, we get:
$$ 18x + 32y \frac{dy}{dx} = 0 $$
$$ 9x + 16y \frac{dy}{dx} = 0 $$
$$ \frac{dy}{dx} = -\frac{9x}{16y} $$

At point $(4,3)$:
$$ \left[\frac{dy}{dx}\right]_{(4,3)} = -\frac{9 \times 4}{16 \times 3} = -\frac{36}{48} = -\frac{3}{4} $$

The equation of the tangent at $(4,3)$ is:
$$ y - 3 = -\frac{3}{4}(x - 4) $$
$$ 4(y - 3) = -3(x - 4) $$
$$ 4y - 12 = -3x + 12 $$
$$ 3x + 4y = 24 $$

Answer: $3x + 4y = 24$

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7. The circle $x^2+y^2-4x-6y-3=0$ at $(2,-1)$.

Solution:

The equation of the circle is:
$$ x^2 + y^2 - 4x - 6y - 3 = 0 \quad \dots(1) $$

Differentiating both sides of (1) w.r.t. $x$, we get:
$$ 2x + 2y \frac{dy}{dx} - 4 - 6 \frac{dy}{dx} = 0 $$
$$ 2(x - 2) + 2 \frac{dy}{dx}(y - 3) = 0 $$
$$ (x - 2) + (y - 3) \frac{dy}{dx} = 0 $$
$$ \frac{dy}{dx} = -\frac{x - 2}{y - 3} = \frac{2 - x}{y - 3} $$

At point $(2,-1)$:
$$ \left[\frac{dy}{dx}\right]_{(2,-1)} = \frac{2 - 2}{-1 - 3} = \frac{0}{-4} = 0 $$

The equation of the tangent at $(2,-1)$ is:
$$ y - (-1) = 0 \cdot (x - 2) $$
$$ y + 1 = 0 $$

Answer: $y + 1 = 0$

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8. The rectangular hyperbola $xy=16$ at $(-4,-4)$.

Solution:

The equation of the curve is:
$$ xy = 16 \quad \dots(1) $$

Differentiating both sides of (1) w.r.t. $x$, we get:
$$ x \frac{dy}{dx} + y \cdot 1 = 0 $$
$$ \frac{dy}{dx} = -\frac{y}{x} $$

At point $(-4,-4)$:
$$ \left[\frac{dy}{dx}\right]_{(-4,-4)} = -\frac{-4}{-4} = -1 $$

The equation of the tangent at $(-4,-4)$ is:
$$ y - (-4) = -1 (x - (-4)) $$
$$ y + 4 = -(x + 4) $$
$$ y + 4 = -x - 4 $$
$$ x + y + 8 = 0 $$

Answer: $x + y + 8 = 0$

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9. The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $(x,y)$.

Solution:

Rewriting the equation:
$$ b^2x^2 - a^2y^2 = a^2b^2 \quad \dots(1) $$

Differentiating both sides of (1) w.r.t. $x$:
$$ 2b^2x - 2a^2y \frac{dy}{dx} = 0 $$
$$ b^2x = a^2y \frac{dy}{dx} $$
$$ \frac{dy}{dx} = \frac{b^2x}{a^2y} $$

The equation of the tangent at point $(x_1, y_1)$ [using general coordinates $(x,y)$ from question context]:
$$ Y - y = \frac{b^2x}{a^2y} (X - x) $$
$$ a^2y(Y - y) = b^2x(X - x) $$
$$ a^2yY - a^2y^2 = b^2xX - b^2x^2 $$
$$ b^2x^2 - a^2y^2 = b^2xX - a^2yY $$

Substituting $b^2x^2 - a^2y^2 = a^2b^2$ from eq (1):
$$ a^2b^2 = b^2xX - a^2yY $$

Answer: $b^2xX - a^2yY = a^2b^2$

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10. The parabola $y^2=-36x$ at point whose ordinate is three times the abscissa.

Solution:

The equation is:
$$ y^2 = -36x \quad \dots(1) $$

Let the point be $(h, k)$. Given that ordinate ($k$) is three times the abscissa ($h$), so $k = 3h$.
The point $(h, 3h)$ lies on the parabola:
$$ (3h)^2 = -36h $$
$$ 9h^2 = -36h $$
Since $h \neq 0$, dividing by $9h$: $$ h = -4 $$
Therefore, $k = 3(-4) = -12$.
The point is $(-4, -12)$.

Differentiating (1) w.r.t. $x$:
$$ 2y \frac{dy}{dx} = -36 \implies \frac{dy}{dx} = \frac{-18}{y} $$

Slope at $(-4, -12)$:
$$ m = \left[\frac{dy}{dx}\right]_{(-4,-12)} = \frac{-18}{-12} = \frac{3}{2} $$

Equation of tangent:
$$ y - (-12) = \frac{3}{2} (x - (-4)) $$
$$ y + 12 = \frac{3}{2} (x + 4) $$
$$ 2(y + 12) = 3(x + 4) $$
$$ 2y + 24 = 3x + 12 $$
$$ 3x - 2y - 12 = 0 \implies 3x - 2y = 12 $$

Answer: $3x - 2y = 12$

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11. The ellipse $x^2+4y^2=25$ at point whose ordinate is $2$.

Solution:

The equation is:
$$ x^2 + 4y^2 = 25 \quad \dots(1) $$

Given ordinate $y = 2$. Substitute into (1):
$$ x^2 + 4(2)^2 = 25 $$
$$ x^2 + 16 = 25 \implies x^2 = 9 \implies x = \pm 3 $$

The points are $(3, 2)$ and $(-3, 2)$.

Differentiating (1) w.r.t. $x$:
$$ 2x + 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y} $$

Case 1: At $(3, 2)$
$$ m = \left[\frac{dy}{dx}\right]_{(3,2)} = -\frac{3}{4(2)} = -\frac{3}{8} $$
Equation of tangent:
$$ y - 2 = -\frac{3}{8}(x - 3) $$
$$ 8(y - 2) = -3(x - 3) $$
$$ 8y - 16 = -3x + 9 $$
$$ 3x + 8y = 25 $$

Case 2: At $(-3, 2)$
$$ m = \left[\frac{dy}{dx}\right]_{(-3,2)} = -\frac{-3}{4(2)} = \frac{3}{8} $$
Equation of tangent:
$$ y - 2 = \frac{3}{8}(x + 3) $$
$$ 8(y - 2) = 3(x + 3) $$
$$ 8y - 16 = 3x + 9 $$
$$ 3x - 8y + 25 = 0 $$

Answers: $3x + 8y = 25$ and $3x - 8y + 25 = 0$

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12. The circle $x^2+y^2-6x-2y+6=0$ at points equidistant from coordinate axes.

Solution:

The equation is:
$$ x^2 + y^2 - 6x - 2y + 6 = 0 \quad \dots(1) $$

Points equidistant from axes satisfy $y = x$ (first/third quadrant logic usually implies magnitude, but here we check intersection with $y=x$). Let the point be $(a, a)$.

Substitute $(a,a)$ into (1):
$$ a^2 + a^2 - 6a - 2a + 6 = 0 $$
$$ 2a^2 - 8a + 6 = 0 $$
$$ a^2 - 4a + 3 = 0 $$
$$ (a - 3)(a - 1) = 0 \implies a = 3, 1 $$

The points are $(3, 3)$ and $(1, 1)$.

Differentiating (1) w.r.t. $x$:
$$ 2x + 2y \frac{dy}{dx} - 6 - 2 \frac{dy}{dx} = 0 $$
$$ 2(x - 3) + 2 \frac{dy}{dx}(y - 1) = 0 $$
$$ \frac{dy}{dx} = -\frac{x - 3}{y - 1} = \frac{3 - x}{y - 1} $$

At $(3, 3)$:
$$ m = \frac{3 - 3}{3 - 1} = 0 $$
Tangent is horizontal:
$$ y - 3 = 0(x - 3) \implies y = 3 $$

At $(1, 1)$:
Note: The derivative formula gives division by zero if applied directly blindly, but looking at the implicit differentiation step: $(y-1)y' = -(x-3)$. If $y=1$, then $0 = -(x-3) \implies x=3$. But our point is $(1,1)$. Let's re-evaluate slope at $(1,1)$.
Actually, substituting $(1,1)$ into $\frac{dy}{dx} = \frac{3-x}{y-1}$ results in $\frac{2}{0}$, which implies an infinite slope (vertical tangent).
Alternatively, swap variables for vertical line equation: $x = constant$.
Equation of tangent at $(1,1)$ is $x = 1$ or $x - 1 = 0$.

Answers: $y - 3 = 0$ and $x - 1 = 0$

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13. The curve $x^{\frac 23}+y^{\frac 23}=a^{\frac 23}$ at $(x_1,y_1)$.

Solution:

The equation is:
$$ x^{2/3} + y^{2/3} = a^{2/3} \quad \dots(1) $$

Differentiating w.r.t. $x$:
$$ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 $$
$$ x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0 $$
$$ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} $$

Slope at $(x_1, y_1)$:
$$ m = -\left(\frac{y_1}{x_1}\right)^{1/3} $$

Equation of tangent:
$$ y - y_1 = -\left(\frac{y_1}{x_1}\right)^{1/3} (x - x_1) $$
Multiply by $x_1^{1/3}$:
$$ x_1^{1/3}(y - y_1) = -y_1^{1/3}(x - x_1) $$
$$ x_1^{1/3}y - x_1^{1/3}y_1 = -xy_1^{1/3} + x_1y_1^{1/3} $$
$$ xy_1^{1/3} + x_1^{1/3}y = x_1y_1^{1/3} + x_1^{1/3}y_1 $$
Factor RHS:
$$ xy_1^{1/3} + x_1^{1/3}y = x_1^{1/3}y_1^{1/3} (x_1^{2/3} + y_1^{2/3}) $$

Since $(x_1, y_1)$ is on the curve, $x_1^{2/3} + y_1^{2/3} = a^{2/3}$.
$$ xy_1^{1/3} + x_1^{1/3}y = x_1^{1/3}y_1^{1/3} a^{2/3} $$

Divide by $x_1^{1/3}y_1^{1/3}$:
$$ \frac{x}{x_1^{1/3}} + \frac{y}{y_1^{1/3}} = a^{2/3} $$
Or written as:
$$ xx_1^{-1/3} + yy_1^{-1/3} = a^{2/3} $$

Answer: $xx_1^{-1/3} + yy_1^{-1/3} = a^{2/3}$

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14. The ellipse $x=a\cos\theta,~y=b\sin\theta$ at $\theta=\frac{\pi}{3}$.

Solution:

Parametric equations:
$$ x = a \cos \theta \implies \frac{dx}{d\theta} = -a \sin \theta $$
$$ y = b \sin \theta \implies \frac{dy}{d\theta} = b \cos \theta $$

Slope:
$$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta $$

At $\theta = \frac{\pi}{3}$:
$$ m = -\frac{b}{a} \cot \left(\frac{\pi}{3}\right) = -\frac{b}{a} \cdot \frac{1}{\sqrt{3}} = -\frac{b}{a\sqrt{3}} $$

Coordinates at $\theta = \frac{\pi}{3}$:
$$ x_1 = a \cos \left(\frac{\pi}{3}\right) = \frac{a}{2} $$
$$ y_1 = b \sin \left(\frac{\pi}{3}\right) = \frac{b\sqrt{3}}{2} $$

Equation of tangent:
$$ y - \frac{b\sqrt{3}}{2} = -\frac{b}{a\sqrt{3}} \left(x - \frac{a}{2}\right) $$
Multiply by $2a\sqrt{3}$ to clear denominators:
$$ 2a\sqrt{3}y - 3ab = -2b \left(x - \frac{a}{2}\right) \cdot \sqrt{3} \cdot \frac{a}{a} \dots \text{Wait, simpler algebra:} $$
$$ y - \frac{b\sqrt{3}}{2} = -\frac{b}{a\sqrt{3}}x + \frac{b}{2\sqrt{3}} $$
Multiply entire equation by $a\sqrt{3}$:
$$ a\sqrt{3}y - \frac{3ab}{2} = -bx + \frac{ab}{2} $$
$$ bx + a\sqrt{3}y = \frac{3ab}{2} + \frac{ab}{2} $$
$$ bx + a\sqrt{3}y = 2ab $$

Dividing by $ab$:
$$ \frac{x}{a} + \frac{\sqrt{3}y}{b} = 2 $$

Answer: $\frac{x}{a} + \frac{\sqrt{3}y}{b} = 2$

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15. The curve $x^3+xy^2-3x^2+4x+5y+2=0$ at $(1,-1)$.

Solution:

The equation is:
$$ x^3 + xy^2 - 3x^2 + 4x + 5y + 2 = 0 \quad \dots(1) $$

Differentiating w.r.t. $x$:
$$ 3x^2 + (1 \cdot y^2 + x \cdot 2y \frac{dy}{dx}) - 6x + 4 + 5 \frac{dy}{dx} = 0 $$
Grouping $\frac{dy}{dx}$ terms:
$$ \frac{dy}{dx} (2xy + 5) = 6x - 3x^2 - y^2 - 4 $$
$$ \frac{dy}{dx} = \frac{-3x^2 - y^2 + 6x - 4}{2xy + 5} $$

At point $(1, -1)$:
$$ \text{Numerator} = -3(1)^2 - (-1)^2 + 6(1) - 4 = -3 - 1 + 6 - 4 = -2 $$
$$ \text{Denominator} = 2(1)(-1) + 5 = -2 + 5 = 3 $$
$$ m = \left[\frac{dy}{dx}\right]_{(1,-1)} = -\frac{2}{3} $$

Equation of tangent:
$$ y - (-1) = -\frac{2}{3} (x - 1) $$
$$ y + 1 = -\frac{2}{3} (x - 1) $$
$$ 3(y + 1) = -2(x - 1) $$
$$ 3y + 3 = -2x + 2 $$
$$ 2x + 3y + 1 = 0 $$

Answer: $2x + 3y + 1 = 0$