Question

If x + y = k is normal to the parabola $y^2=12x,p$ is the length of perpendicular from the focus of the parabola on this normal, then $3k^3+2p^2$ is equal to

• A. 2223
• B. 2224
• C. 2222
• D. None of these

Answer 1

Answer: (A)

2223

We have,

$y^2=12x.......\left(1\right)$

On comparing that,

$y^2=4ax$

$a=3$

Then the focus is $F(a,0)=(3,0)$

On differentiating and we get,

$2y\frac{dy}{dx}=12$

$\frac{dy}{dx}=\frac{6}{y}$

At the point

$(x_1,y_1)$

${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{6}{y_1}$

But given that,

$x+y=k$ is the normal.

Then,

Slope of normal

$=-1$

$\frac{6}{y_1}\times (-1)=-1$

$y_1=6$

But also,

${y_1}^2=12x_1$

$6^2=12x_1$

$x_1=36$

But given equation

$x_1+y_1=k$

$3+6=k$

$k=9$

Now. Equation of normal is $x+y=9$

Perpendicular from focus onto the normal.

$p=\left|\frac{x_1+y_1-9}{\sqrt{1+1}}\right|$

$p=\left|\frac{3+0-9}{\sqrt{1+1}}\right|$

$p=\frac{6}{\sqrt{2}}$

Now,

$3k^3+2p^2$

$=3\times 9^3+2\times \frac{36}{2}$

$=2187+36$

$=2223$

Hence, this is the answer.