Question:
If the line \(y = 4x-5\) is tangent to the curve \(y^2 = ax^3+b\) at the point \((2,3)\), then the value of \(7a-2b\) is...
If the line \(y = 4x-5\) is tangent to the curve \(y^2 = ax^3+b\) at the point \((2,3)\), then the value of \(7a-2b\) is...
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Equal intercepts means the normal has slope -1.
Updated On: Oct 1, 2026
- \(0\)
- \(7\)
- \(14\)
- \(28\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
A line making equal intercepts on the axes has slope -1. So the normal to the curve at the required point has slope -1, and the tangent has slope +1.
Step 2: Slope of the tangent:
Differentiate \(9y^2 = x^3\): \(18y\,y' = 3x^2\), so \(y' = \frac{x^2}{6y}\). Set \(y' = 1\): \(x^2 = 6y\), that is \(y = \frac{x^2}{6}\).
Step 3: Use the curve equation:
\[ 9\left(\frac{x^2}{6}\right)^2 = x^3 \Rightarrow \frac{x^4}{4} = x^3 \Rightarrow x = 4\ (x \ne 0) \]
Then \(y = \frac{16}{6} = \frac83\). The point is \(\left(4, \frac83\right)\).
The points with \(x = -4\) do not lie on the curve, since \(x^3 = -64\) cannot equal \(9y^2 \ge 0\).
Final Answer:
The point is \(\left(4, \frac83\right)\), option (B). \[ \boxed{\left(4,\frac{8}{3}\right)} \]
A line making equal intercepts on the axes has slope -1. So the normal to the curve at the required point has slope -1, and the tangent has slope +1.
Step 2: Slope of the tangent:
Differentiate \(9y^2 = x^3\): \(18y\,y' = 3x^2\), so \(y' = \frac{x^2}{6y}\). Set \(y' = 1\): \(x^2 = 6y\), that is \(y = \frac{x^2}{6}\).
Step 3: Use the curve equation:
\[ 9\left(\frac{x^2}{6}\right)^2 = x^3 \Rightarrow \frac{x^4}{4} = x^3 \Rightarrow x = 4\ (x \ne 0) \]
Then \(y = \frac{16}{6} = \frac83\). The point is \(\left(4, \frac83\right)\).
The points with \(x = -4\) do not lie on the curve, since \(x^3 = -64\) cannot equal \(9y^2 \ge 0\).
Final Answer:
The point is \(\left(4, \frac83\right)\), option (B). \[ \boxed{\left(4,\frac{8}{3}\right)} \]
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