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If $l x+m y+n=0$ is tangent to the parabola $x^2=y$, then condition of tangency is
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Verified Answer
The correct answer is:
$I^2=4 m n$
Given that $l x+m y+n+0 \ldots(i)$
$x^2=y \ldots(ii)$
The point of intersection of the line and parabola are obtained by solving (i) and (ii) simultaneously
substituting the values of $x$ from (i) in (ii), we get $\left(\frac{m y+n}{l}\right)^2=y \quad \Rightarrow m^2 y^2+n^2+2 m n y=y l^2$
$\Rightarrow m^2 y^2+\left(2 m n-l^2\right) y+n^2=0 \ldots(iii)$
If lines (iii) touches the parabola (ii), then discriminant
$=0 \Rightarrow\left(2 m n-l^2\right)^2=4 m^2 n^2$
$\Rightarrow 4 m^2 n^2+l^4-4 m n l^2=4 m^2 n^2 \Rightarrow l^2=4 m n$.
$x^2=y \ldots(ii)$
The point of intersection of the line and parabola are obtained by solving (i) and (ii) simultaneously
substituting the values of $x$ from (i) in (ii), we get $\left(\frac{m y+n}{l}\right)^2=y \quad \Rightarrow m^2 y^2+n^2+2 m n y=y l^2$
$\Rightarrow m^2 y^2+\left(2 m n-l^2\right) y+n^2=0 \ldots(iii)$
If lines (iii) touches the parabola (ii), then discriminant
$=0 \Rightarrow\left(2 m n-l^2\right)^2=4 m^2 n^2$
$\Rightarrow 4 m^2 n^2+l^4-4 m n l^2=4 m^2 n^2 \Rightarrow l^2=4 m n$.
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