AP EAMCET · Maths · Ellipse
If \(4 x-3 y-5=0\) is a normal to the ellipse \(3 x^2+8 y^2=k\), then the equation of the tangent drawn to this ellipse at the point \((-2, m)(m\gt0)\) is
- A \(3 x+4 y-14=0\)
- B \(3 x-4 y+10=0\)
- C \(3 x-4 y+1=0\)
- D \(4 x+3 y-3=0\)
Answer & Solution
Correct Answer
(B) \(3 x-4 y+10=0\)
Step-by-step Solution
Detailed explanation
Given the equation of ellipse \(3 x^2+8 y^2=k\) \(\Rightarrow \frac{x^2}{\frac{k}{3}}+\frac{y^2}{\frac{k}{8}}=1\) Now, equation of normal to the ellipse at \(\left(x_1, y_1\right)\) is \(\frac{a^2}{x_1} x-\frac{b^2}{y_1} y=a^2-b^2\)…
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