Question

The condition that the line $\frac{x}{p}+\frac{y}{q}=1$ to be tangent to line $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

• A. $\frac{a^2}{p^2}+\frac{b^2}{q^2}=1$
• B. $\frac{a^2}{p^2}-\frac{b^2}{q^2}=1$
• C. $\frac{a^2}{q^2}-\frac{b^2}{p^2}=1$
• D. $\frac{a^2}{q^2}+\frac{b^2}{p^2}=1$

Answer 1

Answer: (B)

$\frac{a^2}{p^2}-\frac{b^2}{q^2}=1$

Equation of tangents to at $(x,y)$ is

$\frac{x^2}{y^2}-\frac{y^2}{b^2}=1$

$\frac{xx1}{a^2}-\frac{yy1}{b^2}=1$

now, if $\frac{x}{p}+\frac{y}{q}=1$ is a tangents at $(x,)$

then

$\frac{x1}{a^2}=\frac{-y1}{b^2}=\frac{1}{1}$

$=\frac{x1}{a^2}p=\frac{-y1}{b^2}q=1$

$x_1=\frac{a^2}{p}andy1=\frac{-b^2}{q}$

$Since\left(x_1,y_1\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$so,\frac{x{1}^2}{a^2}-\frac{y{1}^2}{b^2}=1$

$=\frac{a^4}{a^2\times b^2}-\frac{b^4}{q^2\times b^2}=1$

$\frac{a^2}{p^2}-\frac{b^2}{q^2}=1$