Question

The x-intercept of the tangent at any arbitrary point of the curve $\frac{a}{x^2}+\frac{b}{y^2}=1$ is proportional to:

• A. square of the abscissa of the point of tangency
• B. square root of the abscissa of the point of tangency
• C. cube of the abscissa of the point of tangency
• D. cube root of the abscissa of the point of tangency

Answer 1

Answer: (C)

cube of the abscissa of the point of tangency

let $P(\frac{\sqrt{a}}{\cos \theta },\frac{\sqrt{b}}{\sin \theta })$ be any point of curve $\frac{a}{x^2}+\frac{b}{y^2}=1$
$\frac{-2a}{x^3}-\frac{2b}{y^3}\frac{dy}{dx}=0$
$\frac{dy}{dx}$at point P$=-\sqrt{\frac{b}{a}}{\cot }^3\theta$
Equation of tangent at point P: $y-\frac{\sqrt{b}}{\sin \theta }=-\sqrt{\frac{b}{a}}{\cot }^3\theta (x-\frac{\sqrt{a}}{\cos \theta })$
for X-intercept put $x=0$
$0-\frac{\sqrt{b}}{\sin \theta }=-\sqrt{\frac{b}{a}}{\cot }^3\theta (x-\frac{\sqrt{a}}{\cos \theta })$
$\Rightarrow x=\sqrt{a}(\frac{{\sin }^2\theta }{{\cos }^3\theta }+\frac{1}{\cos \theta })=\frac{\sqrt{a}}{{\cos }^3\theta }$
Therefore, X-intercept is proportional to cube of abcissa.
Ans: C