Question

If the normal drawn to the curve \(y^4=16x^3\) at the point of intersection of this curve and the line \(y=2\) meets the \(X\)- and \(Y\)-axes at \(A\) and \(B\) respectively, then \(OA+3OB=\)

• A. \(6\)
• B. \(8\)
• C. \(16\)
• D. \(12\)

Hint:

For normals: \[ m_n=-\frac1{m_t} \] Always compute the tangent slope first through differentiation, then immediately take the negative reciprocal to obtain the normal slope.

Answer 1

The Correct Option is D

Step 1: Find the point on the curve.
Put $y=2$ into $y^4=16x^3$: $16=16x^3$, so $x^3=1$ and $x=1$. The point is $(1,2)$.
Step 2: Get the tangent slope by implicit differentiation.
From $y^4=16x^3$, $4y^3\dfrac{dy}{dx}=48x^2$, so $\dfrac{dy}{dx}=\dfrac{12x^2}{y^3}$.
Step 3: Evaluate the slope at $(1,2)$.
$m_t=\dfrac{12(1)}{8}=\dfrac32$. The normal slope is $m_n=-\dfrac{1}{m_t}=-\dfrac23$.
Step 4: Write the normal line.
$y-2=-\dfrac23(x-1)$. Multiply by $3$: $3y-6=-2x+2$, so $2x+3y-8=0$.
Step 5: Find the intercepts $A$ and $B$.
Put $y=0$: $2x=8$, $x=4$, so $OA=4$. Put $x=0$: $3y=8$, $y=\dfrac83$, so $OB=\dfrac83$.
Step 6: Compute $OA+3OB$.
\[ OA+3OB=4+3\cdot\frac83=4+8=12. \] \[ \boxed{12} \]