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  • The equations of the sides AB and AC of a triangle ABC are <img alt="(\lambda+1) x+\lambda y=4" src="https://entrancecorner.oncodecogs.com/gif.latex?%28%5Clambda&plus;1%29%20x&plus;%5C

The equations of the sides AB and AC of a triangle ABC are (\lambda+1) x+\lambda y=4 and \lambda x+(1-\lambda) y+\lambda=0respectively. 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 : 

Option: 1

4


Option: 2

2


Option: 3

\sqrt{6}  


Option: 4

2\sqrt{2}


Answers (1)

best_answer

\begin{aligned} & (\lambda+1)(1-\lambda) x+\lambda(1-\lambda) y=4(1-\lambda) \\ & \lambda^2 x+\lambda(1-\lambda) \mathrm{y}=-\lambda^2 \\ & -\quad-\quad+ \\ & \left(1-\lambda^2-\lambda^2\right) x=4-4 \lambda+\lambda^2 \\ & \left(1-2 \lambda^2\right) \mathrm{x}=4-4 \lambda+\lambda^2 \\ & \end{aligned}

x=0 \Rightarrow \lambda=2

\left.\begin{array}{l} \mathrm{AB}: 3 \mathrm{x}+2 \mathrm{y}=4 \\ \mathrm{AC}: 2 \mathrm{x}-\mathrm{y}+2=0 \end{array}\right\} \mathrm{A}(0,2)

\mathrm{CH} \perp \mathrm{AB}

\begin{array}{rlrl} \left(\frac{b-2}{a-1}\right) \times\left(\frac{-3}{2}\right) & =-1 & \\ & 3 b-6=2 a-2 & & \text { Also } 2 a-b+2=0 \\ & 3 b-2 a=4 & & b=2 a+2 \end{array}

\begin{aligned} & 6 a+6=2 a+4 \quad C\left(-\frac{1}{2}, 1\right) \\ & 4 a=-2 \\ & a=\frac{-1}{2}, b=1 \\ & \therefore y^2=6 x \end{aligned}

\begin{aligned} & t y=x+\frac{3}{2} t^2 \\ & t=-\frac{1}{2}+\frac{3}{2} t^2 \\ & 3 t^2-1=2 t \\ & 3 t^2-2 t-1=0 \\ & 3 t^2-3 t+t-1=0 \\ & (3 t+1)(t-1)=0 \end{aligned}

\begin{aligned} P\left(\frac{3}{2}, 3\right) \quad & \therefore d=\sqrt{\left(\frac{3}{2}+\frac{1}{2}\right)^2+(3-1)^2} \\ & =\sqrt{4+4}=2 \sqrt{2} \end{aligned}

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rishi.raj

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