Let L₁ be a tangent to the parabola y² = 4(x + 1)
and L₂ be a tangent to the parabola y² = 8(x + 2)
such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line :

L₁ : y² = 4(x + 1) <br><br>Equation of tangent y = m(x + 1) + ${1 \over m}$ ...(1) <br><br>L₂ : y² = 8(x + 2) <br><br>Equation of tangent y = m'(x + 2) + ${2 \over {m'}}$ <br><br>$\Rightarrow$ y = m'x + 2$\left( {m' + {1 \over {m'}}} \right)$ ....(2) <br><br>Since lines intersect at right angles <br><br>$\therefore$ mm' = -1 <br><br>$\Rightarrow$ m' = ${ - {1 \over {m}}}$ <br><br>Putting it in equation (2) <br><br>y = $- {1 \over m}x + 2\left( { - {1 \over m} - m} \right)$ <br><br>$\Rightarrow$ y = $- {1 \over m}x - 2\left( {m + {1 \over m}} \right)$ ....(3) <br><br>From equation (1) and (3) <br><br>m(x + 1) + ${1 \over m}$ = $- {1 \over m}x - 2\left( {m + {1 \over m}} \right)$ <br><br>$\Rightarrow$ $\left( {m + {1 \over m}} \right)x + 3\left( {m + {1 \over m}} \right)$ = 0 <br><br>$\therefore$ x + 3 = 0