Question:
If the tangents drawn from a point \( P \) to the ellipse \( 4x^2+9y^2-16x+54y+61=0 \) are perpendicular, then the locus of \( P \) is:
If the tangents drawn from a point \( P \) to the ellipse \( 4x^2+9y^2-16x+54y+61=0 \) are perpendicular, then the locus of \( P \) is:
Show Hint
For perpendicular tangents from an external point to an ellipse, the locus of such points satisfies:
\[
x^2 + y^2 - 2gx - 2fy = 0
\]
where \( g, f \) are derived from the ellipse equation.
Updated On: Jun 5, 2025
- \( x^2+y^2-4x+6y+4=0 \)
- \( x^2+y^2-4x+6y=0 \)
- \( x^2+y^2-6x+4y+9=0 \)
- \( x^2+y^2-6x+4y=0 \)
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The Correct Option is D
Solution and Explanation
The given ellipse equation:
\[
4x^2 + 9y^2 - 16x + 54y + 61 = 0
\]
Rewriting in standard form:
\[
\frac{(x-2)^2}{4} + \frac{(y+3)^2}{9} = 1
\]
If tangents from a point \( P(h, k) \) to an ellipse are perpendicular, the locus satisfies:
\[
h^2 + k^2 - 6h + 4k = 0
\]
Thus, the correct answer is:
\[
x^2+y^2-6x+4y=0
\]
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