JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The equations of the sides \(AB\) and \(AC\) of a triangle \(ABC\) are \((\lambda+1) x +\lambda y =4 \text { and } \lambda x +(1-\lambda) y +\lambda=0\) respectively. Its vertex \(A\) is on the \(y\)-axis and its orthocentre is \((1,2)\). The length of the tangent from the point \(C\) to the part of the parabola \(y^2=6 x\) in the first quadrant is
- A \(\sqrt{6}\)
- B \(2 \sqrt{2}\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(AB :(\lambda+1) x +\lambda y =4\) \(AC : \lambda x +(1-\lambda) y +\lambda=0\) Vertex \(A\) is on \(y\)-axis \(\Rightarrow x=0\) So \(y =\frac{4}{\lambda}, y =\frac{\lambda}{\lambda-1}\) \(\Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1}\) \(\Rightarrow \lambda=2\)…
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