Question

If $px+qy+r=0$ is tangent of the parabola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ then

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Ay tangent of slope 'm' to the parabola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

can be given as $y=\pm \sqrt{a^2{m}^2-b^2}$

Since $px+qy+r=0$ is given as the targent then we can

compare the cofficient as below.

$y=-\frac{p}{q}\times -\frac{r}{q}$

$m=-\frac{p}{q}$ - - - - - - -(1)

$-\frac{r}{q}=\pm \sqrt{a^2{m}^2-b^2}$ - - - - - - (2)

i.e.,${(-\frac{r}{q})}^2=a^2.\frac{p^2}{q^2}-b^2$ using (1)

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

$a^2{p}^2-b^2{q}^2=r^2$

hence option (d) is correct