Question

The tangents drawn at the extremeties of a focal chord of the parabola $y^2=16x$.

• A. Intersect on $x=0$
• B. Intersect on the line $x+4=0$
• C. Intersect at an angle of ${60}^{\circ }$
• D. Intersect at an angle of ${45}^{\circ }$

Answer 1

Answer: (B)

Intersect on the line $x+4=0$

Consider the equation of the parabola, $y^2=16x$. Hence the focus is $F=(\frac{16}{4},0)=(4,0)$.
Substituting $x=4$ in the equation of the parabola.
$y^2=64$
$y=\pm 8$.
Hence the points at the extremities of the focal chord are $(4,8)$ and $(4,-8)$
Slope of the tangent at the point $(4,-8)$ implies
$2y.y^{\prime }=16$ or $y^{\prime }=\frac{8}{y}$. Therefore, $m=-1$.
Hence equation of the tangent
$\frac{y+8}{x-4}=-1$ or $y+8=-x+4$ or $x+y=-4$ ... $\left(i\right)$
Similarly
Slope of the tangent at the point $(4,8)$ implies
$2y.y^{\prime }=16$ or $y^{\prime }=\frac{8}{y}$. Therefore, $m=1$.
Hence equation of the tangent
$\frac{y-8}{x-4}=1$ or $y-8=x-4$ or $x-y=-4$ ... $\left(ii\right)$
Therefore solving equation $\left(i\right)$ and $\left(ii\right)$
$x=-4$ and $y=0$.

Hence, they intersect along the line $x=-4$ or $x+4=0$.
The angle between the two tangents is
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$
Now $m_1.m_2=-1$ hence $\tan \theta =\infty$ and $\theta ={90}^0$. Hence the tangents intersect at ${90}^0$.

Hence, they intersect along the line $x+4=0$ with an angle of ${90}^o$.