Question:
If the tangent drawn to the curve \( y = x^3 \) at point \( (\alpha, \beta) \) cuts again the curve at another point \( (\alpha_1, \beta_1) \), then \( \frac{\beta_1}{\beta} = \):
If the tangent drawn to the curve \( y = x^3 \) at point \( (\alpha, \beta) \) cuts again the curve at another point \( (\alpha_1, \beta_1) \), then \( \frac{\beta_1}{\beta} = \):
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Find second intersection using tangent line and original curve equation.
Updated On: May 13, 2025
- \( -2 \)
- \( 1 \)
- \( -8 \)
- \( 27 \)
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The Correct Option is C
Solution and Explanation
Given \( y = x^3 \Rightarrow y' = 3x^2 \)
Equation of tangent at \( x = \alpha \):
\[
y - \alpha^3 = 3\alpha^2(x - \alpha)
\Rightarrow y = 3\alpha^2 x - 2\alpha^3
\]
Set equal to curve again:
\[
x^3 = 3\alpha^2 x - 2\alpha^3 \Rightarrow \text{Solve to find } x = -2\alpha \Rightarrow y = (-2\alpha)^3 = -8\alpha^3
\Rightarrow \frac{\beta_1}{\beta} = \frac{-8\alpha^3}{\alpha^3} = \boxed{-8}
\]
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