1STOP CALLING ME "AVEEK" - I"M THE DARK KNIGHT !!!
Q1 if y=2x-3 is a tangent to parabola y2=4a(x-13) then a= ??
i did this way..eqn of tangent to parabola:y=mx-m/3+a/m
given that y=2x-3 is tangent
Comparing gives us m=2
and proceeding we get a=-14/3
but ans is given a=2±√583
Q2 length of common chord of parabola y^2=4(x+3) and circle x^2+y^2+4x=0 is ??
Q3 a normal to parabola y^2=4ax makes angle φ with x-axis where tanφ=2.if the curve cuts again at angle θ,then θ= ??
eqn of normal will be y=mx-2am-am3
putting m=tanφ=2 we get y=2x-12a
what next ???or am i wrong ??
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11 Answers
1Eureka, ist problem i can elaborate a method but to post the full soln. it will take lot of time. So just showing the method.
Eliminate y from the equations of the parabola and the st. line by squaring y of the tangent equation and equating it to the parabola equation. Now you'll get a quadratic in x. As a tangent touches the parabola in one point only, the discriminant of the quadratic in x =0. Now you will get the value of x. And then you can easily find out the y-co-ordinate and applying it to the equation of the parabola, you will get the value of a.
- Please reply if you r done with it.
The Dark knight
39The method aveek gave is the standard way of solving, discriminant must be zero as the tangent touches at only one point, meaning equal roots. Its the safest method. eure, direct comparison doesn't work as the two equations for tangents you have are NOT completely in the same form. Take LCM as m throughout the RHS, a quadratic in m is obtained. If you had to solve for m, m=2 won't do. Try -m/3 + a/m = -3, but there are two variables so we can't solve by comparison. I used to commit this error quite a lot.
11For the equation of the family of normals....we can say (by comparison) that t=-m....
Now we know t1.....Also we have t_2=-t_1-\frac{2}{t_1} where t_2=sin\theta.....
66Eureka, your method as well as the answer you have got are perfectly okay for Q1. The answer given in the book is not right.
66For the second one, the intersection of the two curves are obtained by the solutions of the equation
x^2 +8x+12=0
i.e. x= -6, -2
But there is no point on either the parabola or on the circle with x=-6. So the parabola and the circle intersects at two points both of which has x=-2 and so y = ±2. So the intersection points are (-2, 2) and (-2, -2) which happens to be the ends of a vertical diameter of the circle. As the the length of the common chord is 4.
1Yes Eurie ... Kaymant Sir is right....either way leads to one answer......but even I get stuck sometimes with this Comparison method......can u please help me out ... Anant Sir ?