JEE Mathematics 2023 Question with Solution

Given: Sides AB and AC are

$$(\lambda+1)x+\lambda y=4$$

and

$$\lambda x+(1-\lambda)y+\lambda=0$$

Vertex A lies on the y-axis and the orthocentre is $(1,2)$.

Find: The length of the tangent from point C to the parabola $y^2=6x$. The solution concludes that the correct option is A.

Since A is on the y-axis, put $x=0$ in both side equations. Then

$$y=\frac{4}{\lambda}, \qquad y=\frac{\lambda-1}{\lambda}$$

Equating these gives

$$\frac{\lambda}{4}=\frac{\lambda-1}{\lambda}$$

so

$$\lambda=2$$

Hence the sides become

$$3x+2y=4$$

and

$$2x-y+2=0$$

Therefore

$$A=(0,2)$$

Let

$$C=(\alpha,2\alpha+2)$$

because C lies on line AC. The slope of AB is $-\frac{3}{2}$, so the altitude through C has slope $\frac{2}{3}$. Since the orthocentre is $(1,2)$, the line joining C and $(1,2)$ must be this altitude. Thus

$$\left(\frac{\alpha-1}{2\alpha}\right)\left(-\frac{2}{3}\right)=-1$$

which gives

$$\alpha=-\frac{1}{2}$$

Hence

$$C=\left(-\frac{1}{2},1\right)$$

For parabola $y^2=4ax$, here

$$4a=6 \Rightarrow a=\frac{3}{2}$$

Let the tangent be

$$y=mx+\frac{2a}{m}=mx+\frac{3}{m}$$

Using the point C as shown in the extracted working, substitution leads to

$$m^2+2m-3=0$$

so

$$m=1,-3$$

The tangent touching the first quadrant corresponds to $m=1$. Then the point of contact is

$$T=\left(\frac{a}{m^2},2a\frac{1}{m}\right)=\left(\frac{3}{2},3\right)$$

The length of the tangent segment is

$$CT=\sqrt{(2)^2+(2)^2}=2\sqrt{2}$$

Therefore, the computed value is $2\sqrt{2}$. However, the solution explicitly marks the correct option as A, even though this value matches option B in the listed choices. Following the solution as primary source, the answer is recorded as A.