Question

If the tangent at a point $P(\alpha ,\beta )$ on the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ cuts the circle $x^2+y^2=25$ at the point $Q(x_1,y_1)$ and $R(x_2,y_2),$ then $\frac{y_1{y}_2}{y_1+y_2}=$

• A. $2\beta$
• B. $\beta$
• C. $\frac{\beta }{2}$
• D. $4\beta$

Answer 1

The correct option is C $\frac{\beta }{2}$

Equation of tangent at $P(\alpha ,\beta )$ on the hyperbola
$\frac{x^2}{25}-\frac{y^2}{16}=1$ is
$\frac{\alpha x}{25}-\frac{\beta y}{16}=1$
$\Rightarrow x=\frac{25}{\alpha }+\frac{25}{16}\cdot \frac{\beta }{\alpha }y\dots \left(1\right)$
As point $P(\alpha ,\beta )$ lies on the hyperbola so,
$\frac{{\alpha }^2}{25}-\frac{{\beta }^2}{16}=1\dots \left(2\right)$

Tangent line $\left(1\right)$ intersects the circle $x^2+y^2=25$
Then
${(\frac{25}{\alpha }+\frac{25}{16}\cdot \frac{\beta }{\alpha }y)}^2+y^2=25$
$\Rightarrow (1+{(\frac{25}{16}\cdot \frac{\beta }{\alpha })}^2)y^2+2\times \frac{25}{\alpha }\times \frac{25\beta }{16\alpha }y+{\left(\frac{25}{\alpha }\right)}^2-25=0$
Let the roots of above quadratic equation are $y_1,y_2.$
$y_1{y}_2=\frac{{\left(\frac{25}{\alpha }\right)}^2-25}{(1+{(\frac{25}{16}\cdot \frac{\beta }{\alpha })}^2)}\dots \left(3\right)$
$y_1+y_2=-\frac{2\times \frac{25}{\alpha }\times \frac{25\beta }{16\alpha }}{(1+{(\frac{25}{16}\cdot \frac{\beta }{\alpha })}^2)}\dots \left(4\right)$
By equation $\left(3\right)$ and $\left(4\right)$
$\frac{y_1{y}_2}{y_1+y_2}=\frac{{\left(\frac{25}{\alpha }\right)}^2-25}{-2\times \frac{25}{\alpha }\times \frac{25\beta }{16\alpha }}$
$=\frac{{\left(\frac{25}{\alpha }\right)}^2[1-\frac{{\alpha }^2}{25}]}{-2\times {\left(\frac{25}{\alpha }\right)}^2\times \frac{\beta }{16}}$
By equation $\left(2\right)$
$\frac{y_1{y}_2}{y_1+y_2}=\frac{-\frac{{\beta }^2}{16}}{-2\times \frac{\beta }{16}}$
$\Rightarrow \frac{y_1{y}_2}{y_1+y_2}=\frac{\beta }{2}$