- #1
utkarshakash
Gold Member
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Homework Statement
Locus of the point of intersection of tangents to the parabolas y[itex]^{2}[/itex]=4(x+1) and y[itex]^{2}[/itex]=8(x+2) which are at right angles, is
Homework Equations
Equation of tangent for first parabola
t[itex]_{1}[/itex]y=x+1+at[itex]_{1}[/itex][itex]^{2}[/itex]
Equation of tangent for second parabola
t[itex]_{2}[/itex]y=x+2+bt[itex]_{2}[/itex][itex]^{2}[/itex]
The Attempt at a Solution
Let us assume that the point of intersection of tangents is (h,k)
Since it lies on the tangent
∴kt[itex]_{1}[/itex]=h+1+at[itex]_{1}[/itex][itex]^{2}[/itex]
[itex]\Rightarrow[/itex]at[itex]_{1}[/itex][itex]^{2}[/itex]-kt[itex]_{1}[/itex]+(h+1)=0
t[itex]_{1}[/itex]t[itex]_{2}[/itex]=h+1 (product of roots)
Also t[itex]_{1}[/itex]t[itex]_{2}[/itex]=-1 (Since they are at right angles)
∴required locus = x+2=0
I can't understand where I'm wrong. Is the equation of tangent incorrect? Please Help.