Question

A pair of tangents with inclinations $\alpha ,\beta$with positive x-axis drawn from an external point to the parabola $y^2=16x$.

If the point P varies such a way that $ta{n}^2\alpha +ta{n}^2\beta =4$ then the locus of p is conic ecentricity.

Answer 1

$y^2=16x$

$\to a=4$

$E{q}^n$ of tangent $y=mx+a/m$

$\Rightarrow y=mx+4/m$

$\Rightarrow my=m^{2}x+4$

$\Rightarrow m^{2}x-my+4=0$

$\Rightarrow m=\frac{+y1\sqrt{y^2-16x}}{2x}$

$m_1^2+m_{\alpha }^2=(m_1+m_2{)}^2-2m_1{m}_2$

$4=\frac{y^2}{x^2}-\alpha \times \frac{4}{x}$

$\Rightarrow 4x^2=y^2-\delta x$

$\Rightarrow 4x^2-8x=y^2$

$\Rightarrow 4(x^2-2x)=y^2$

$\Rightarrow 4\left(\right(x-1{)}^2-1)=y^2$

$\Rightarrow (x-1{)}^2-4-y^2=0$

$\Rightarrow (x-1{)}^2-y^2=4$

$\Rightarrow \frac{(x-1{)}^2}{4}-\frac{y^2}{4}=0$

This is an $e{q}^n$ of hyperbola

$\Rightarrow b^2=a^2(e^2-1)$

$\Rightarrow 1=e^2-1$

$\Rightarrow e^2=2$

$\Rightarrow e=\sqrt{2}$