Question

The tangent at any point P of a curve C meets the x - axis at Q whose abscissa is positive and OP = OQ O being the origin, if C is a family of parabolas having vertix $(\alpha ,\beta )$ and latus rectum = 4a, then evaluate $\frac{4(\alpha +\beta )}{a}$

Answer 1

Equation of tangent
Y- y = m(x - x) $(wherem=\frac{dy}{dx})$
OP =OQ
$\sqrt{x^2+y^2}=x-\frac{y}{m}$
$x^2+y^2={[x-y.\frac{dx}{dy}]}^2$
$\Rightarrow \frac{-ydx+xdy}{\sqrt{x^2+y^2}}=dy$
$\frac{ydx-xdy}{y^2\sqrt{1+\frac{x^2}{y^2}}}=-\frac{1}{y}$
$\frac{1}{\sqrt{1+\frac{x^2}{y^2}}}d\left[\frac{x}{y}\right]=-\frac{1}{y}dy$
On integrating we obtain $\frac{x}{y}+\sqrt{1+\frac{x^2}{y^2}}=\frac{c}{y}$
$y^2=-2c[x-\frac{c}{2}]$
$vertix[\frac{c}{2},0]$
L.R. = 2c = 4a
$\frac{4(\alpha +\beta )}{a}=\frac{4[\frac{c}{2},0]}{\frac{c}{2}}=4$