Question

Three normal are drawn from the point $(14,7)$ to the parabola $y^2-16x-8y=0$. The coordinates of the feet of the normals are

• A. $(0,0),(8,-16),(3,-4)$
• B. $(0,0),(8,16),(3,-4)$
• C. $(0,0),(-8,16),(3,-4)$
• D. $(0,0),(-8,-16),(3,-4)$

Answer 1

The correct option is B $(0,0),(8,16),(3,-4)$

$y^2-16x-8y=0------\left(1\right)$

let P=(14,7), then equation (1) becomes

$y^2-8y+16=16x+16----\left(2\right)$

which is of the form ${(y-l)}^2=4a(x-5)$

$a=4,s=-1,l=4$

let $(-1+4t^2,4+8t)$ be any point on the curve (2)

then from(2), we have $2(y-4)\frac{dy}{dx}=16$

hence $\frac{dy}{dx}=\frac{8}{y-4}$ at $(-1+4t^2,4+8t),\frac{dy}{dx}=\frac{1}{t}$

Hence, the equation of normal at $(-1+4t^2,4+8t)$ is

$y-4-8t=-t(x+1-4t^2)$

$=tx+y-4-8t+t-4t^3=0$

If line passes through the point P(14,7), then

$14t+7-4t^3-7t-4=0$

$4t^3-7t-3=0$

$(t-1)(4t^2-4t-3)=0$

hence, $t=-1,\frac{(4\pm 8)}{8}=-1,\frac{3}{2},\frac{-1}{2}$

when, t=-1, foot of the normal is (3,-4)

when $t=\frac{3}{2}$, foot of the normal is (8,16)

when$t=\frac{1}{2}$ , foot of the normal is (0,0)

So, option (B) is correct.