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 \(OA+3OB=\)
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 \(OA+3OB=\)
Show 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.
Updated On: Jun 17, 2026
- \(6\)
- \(8\)
- \(16\)
- \(12\)
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The Correct Option is D
Solution and Explanation
Concept:
To determine the equation of a normal to a curve:
• First differentiate the equation to obtain the slope of the tangent.
• The slope of the normal is the negative reciprocal of the tangent slope.
• Use point-slope form to obtain the normal equation.
• Finally determine the intercepts on the coordinate axes.
Step 1: Find the point of intersection of the curve and the line \(y=2\). The curve is: \[ y^4 = 16x^3 \] Substitute \(y=2\): \[ 2^4 = 16x^3 \] \[ 16 = 16x^3 \] \[ x^3 = 1 \] \[ x=1 \] Thus, the point is: \[ (1,2) \]
Step 2: Find the slope of the tangent. Differentiate implicitly: \[ y^4 = 16x^3 \] Differentiating both sides: \[ 4y^3\frac{dy}{dx} = 48x^2 \] Therefore, \[ \frac{dy}{dx} = \frac{48x^2}{4y^3} \] \[ \frac{dy}{dx} = \frac{12x^2}{y^3} \] At the point \((1,2)\), \[ m_t = \frac{12(1)^2}{2^3} \] \[ m_t = \frac{12}{8} \] \[ m_t = \frac32 \]
Step 3: Find the slope of the normal. The slope of the normal is: \[ m_n = -\frac{1}{m_t} \] \[ m_n = -\frac{2}{3} \]
Step 4: Find the equation of the normal. Using point-slope form: \[ y-2 = -\frac23(x-1) \] Multiply throughout by \(3\): \[ 3y-6 = -2x+2 \] \[ 2x+3y-8 = 0 \]
Step 5: Find intercepts on axes. For \(X\)-axis intercept, put \(y=0\): \[ 2x-8=0 \] \[ x=4 \] Thus, \[ OA=4 \] For \(Y\)-axis intercept, put \(x=0\): \[ 3y-8=0 \] \[ y=\frac83 \] Thus, \[ OB=\frac83 \]
Step 6: Compute \(OA+3OB\). \[ OA+3OB =4+3\left(\frac83\right) \] \[ =4+8 \] \[ =12 \] Hence, \[ \boxed{12} \]
• First differentiate the equation to obtain the slope of the tangent.
• The slope of the normal is the negative reciprocal of the tangent slope.
• Use point-slope form to obtain the normal equation.
• Finally determine the intercepts on the coordinate axes.
Step 1: Find the point of intersection of the curve and the line \(y=2\). The curve is: \[ y^4 = 16x^3 \] Substitute \(y=2\): \[ 2^4 = 16x^3 \] \[ 16 = 16x^3 \] \[ x^3 = 1 \] \[ x=1 \] Thus, the point is: \[ (1,2) \]
Step 2: Find the slope of the tangent. Differentiate implicitly: \[ y^4 = 16x^3 \] Differentiating both sides: \[ 4y^3\frac{dy}{dx} = 48x^2 \] Therefore, \[ \frac{dy}{dx} = \frac{48x^2}{4y^3} \] \[ \frac{dy}{dx} = \frac{12x^2}{y^3} \] At the point \((1,2)\), \[ m_t = \frac{12(1)^2}{2^3} \] \[ m_t = \frac{12}{8} \] \[ m_t = \frac32 \]
Step 3: Find the slope of the normal. The slope of the normal is: \[ m_n = -\frac{1}{m_t} \] \[ m_n = -\frac{2}{3} \]
Step 4: Find the equation of the normal. Using point-slope form: \[ y-2 = -\frac23(x-1) \] Multiply throughout by \(3\): \[ 3y-6 = -2x+2 \] \[ 2x+3y-8 = 0 \]
Step 5: Find intercepts on axes. For \(X\)-axis intercept, put \(y=0\): \[ 2x-8=0 \] \[ x=4 \] Thus, \[ OA=4 \] For \(Y\)-axis intercept, put \(x=0\): \[ 3y-8=0 \] \[ y=\frac83 \] Thus, \[ OB=\frac83 \]
Step 6: Compute \(OA+3OB\). \[ OA+3OB =4+3\left(\frac83\right) \] \[ =4+8 \] \[ =12 \] Hence, \[ \boxed{12} \]
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