Question

The equation of the normal at the point (a,b) to the curve $y=4x^2$ is x + 24y - 864 = 0. Then, a = and b =

• A. 3
• B. 6
• C. 36
• D. 144

Answer 1

Answer: (A), (C)

The correct options are
A 3
C 36

The curve given is $y=4x^2$

The point on the curve is (a,b). Thus, $b=4a^2$

Also, the slope of tangent at (a,b) is $\frac{dy}{dx}{|}_{(a,b)}$

$\frac{dy}{dx}=\frac{d}{dx}\left(4x^2\right)=4\times 2x=8x$

Thus, $\frac{dy}{dx}{|}_{(a,b)}=8a$

Since the normal is perpendicular to the tangent at the point, the product of its slope and the slope of the tangent is equal to -1.

If m is the slope of the normal, then
$m\times 8a=-1$
$\Rightarrow m=-\frac{1}{8a}$

The normal passes through the point $(a,4a^2)$

Thus, its equation is $y-4a^2=-\frac{1}{8a}(x-a)$

$\Rightarrow 8ay-32a^3+x-a=0$
$\Rightarrow x+8ay-(a+32a^3)=0$

Now, the equation given is x + 24y - 864 = 0

Comparing the coefficients of y, we get 8a = 24
$\Rightarrow a=\frac{24}{8}=3$

$b=4\times 3^2=4\times 9=36$