InterviewSolution
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InterviewSolution
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Consider the parabola whose focus is at (0,0) and tangent at vertex is x-y+1=0 Tangents drawn to the parabola at the extremities of the chord 3x+2y=0 intersect at angle |
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Answer» `pi//6`  The distance between the focus and tangent at the vertex is `(|0-0+1|)/(sqrt(1^(2)+1^(2)))=(1)/(sqrt(2))` The directrix is the line parallel to the tangent vertex ant at a distance `2xx1sqrt(2)` from the focus. Let the EQUATION of the directrix be `x-y+lamda=0` where `(lamda)/(sqrt(1^(2)+1^(2)))=(2)/(sqrt(2))` `:." "lamda=2` Let P (x,y) be any moving point on the parabola. Then, OP=PM `orx^(2)+y^(2)=((x-y+2)/(sqrt(1^(2)+1)))^(2)` `or2x^(2)+=2y^(2)=(x-y+2)^(2)` `orx^(2)+y^(2)+2xy-4x+4y-4=0` Latus rectum length `=2xx` (Distance of focus from directrix) `2|(0-0+2)/(sqrt(1^(2)+1^(2)))|=2sqrt(2)` Solving the parabola with the x-axis, we get `x^(2)-4x-4=0` `orx=(4pmsqrt(32))/(2)=2pm2sqrt(2)` THEREFORE, the length of chord on the x-axis is `4sqrt(2)`. Since the chord 3x+2y=0 PASSES through the focus, it is focus chord. Hence, tangents at the EXTREMITIES of chord are perpendicular. |
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