Tangents Are Drawn From Any Point On The Line X 4a=0 To The Parabola Y^2=4ax. Prove That Chord Of

Let[math] P=(at_1^2,2at_1)[/math] and [math]Q=(at_2^2,2at_2)[/math] be the end points of a chord.Mid point [math]M=(\frac {a(t_1^2+t_2^2) } 2,a(t_1+t_2)) [/math]slope of [math]OP=\dfrac 2 {t_1}[/math] and slope of [math]OQ=\dfrac 2 {t_2} .[/math]Since [math]OP\perp OQ[/math] , [math]\dfrac 2 {t_1} \dfrac 2 {t_2}=-1[/math] or [math]t_1t_2=-4[/math]For[math] M , x=\dfrac {a(t_1^2+t_2^2)} 2=\dfrac {a(t_1+t_2)^2–2at_1t_2} 2=\dfrac {a(t_1+t_2)^2+8a} 2[/math][math]y=a(t_1+t_2) [/math]Eliminating [math] t_1+t_2[/math] we get[math]\boxed{y^2=2a (x-4a)}[/math]

A tangent to the parabola [math]y^2+4bx=0 [/math]meets the parabola [math]y^2=4ax[/math] at P and Q. What is the locus of the midpoint of PQ?00000000000000000000000000000000000Method IEquation of a tangent to the the parabola[math]y^2+4bx=0 \tag 1[/math]is given by[math]y=mx-\dfrac b m \tag 2[/math][math]y^2=4ax \tag 3[/math]Let be represented by parametric equation [math]x=at^2 ; y=2at[/math]Let [math]P[/math] and [math]Q[/math] have parameters[math] t_1[/math] and [math]t_2.[/math]slope of [math]PQ=m=\dfrac {2at_2–2at_1}{a(t_2^2-t_1^2)}=\dfrac {2}{t_2+t_1}[/math]Let M be the midpoint of [math]PQ (h,k)[/math][math]k=a(t_2+t_1)[/math]So [math]m= \dfrac {2a}k[/math]Substituting in [math](1)[/math] and [math](h,k) [/math]lying on the line[math](2a+b)k^2=4a^2h[/math]Replacing[math] (h,k)[/math] by running coordinates [math](x,y)[/math]equation of locus is [math]\boxed{ (2a+b) y^2=4a^2 x}[/math]Method IISome people are allergic to parametric equations.Let [math]x=ny+c[/math] be the equation of the required line.To find intersection point with (1)[math]y^2+4b(ny+c)=0[/math][math]y^2+4bny+4bc=0 \tag 3[/math]Since it is a tangent the two roots of (3) must be equal i.e. discriminant must be 0[math]16b^2n^2–16bc=0 \implies c=bn^2[/math]So the equation of the line becomes[math]x=ny+bn^2 \tag 4[/math]Its intersection with (2) gives[math]y^2=4a(ny+bn^2) or[/math][math]y^2–4any-4abn^2=0[/math]sum of roots[math]=y_1+y_2=4an[/math]Let [math]M(h,k)[/math] be the midpoint of [math]PQ[/math].[math]k=0.5(y_1+y_2)=2an \implies n=\dfrac k{2a}[/math][math](h,k) \text{lies on }(4)[/math][math]h=\dfrac k{2a}k+b(\dfrac k{2a})^2 \implies[/math][math] (2a+b) k^2=4a^2 h[/math]Replacing[math] (h,k)[/math] by running coordinates [math](x,y)[/math][math]\boxed{ (2a+b) y^2=4a^2 x}[/math]

Using parametric equation [math] x=at^2[/math] ; [math]y=2at[/math]substituting in x-y-a =0[math]at^2–2at-a=0[/math] gives t =2 and -1slope=[math]\dfrac {1}{t}[/math] ; [math]m_1= \dfrac{1}{2}[/math] and [math]m_2 =-1[/math][math]\tan \theta = \dfrac {m_1-m_2}{1+m_1m_2}=\dfrac { \dfrac{1}{2}+1}{1+ \dfrac{1}{2}(-1)}=3[/math][math]\theta= \tan^{-1}[/math]3

So I'm guessing that you were given this problem as a homework problem in your JEE preparation, so I will just give you a clue as to how do you find this out. If you still aren't able to do it, message me and I'll send the solution.So, a common tangent to both the curves is a straight line that touches both curves. Note, these two points need not be the same for both these curves (as in the line will touch the circle and parabola at two distinct points).Now, let's say the line that touches the curves isy=mx+c.For the circle, in order to touch the circle, the line would be at a distance equal to the radius from the centre of the circle. That should give you one condition between m and c.Note:- You can also solve this using the condition that since there exists only one root between circle and line; when we solve the equations with each other, the determinant of the quadratic that finally forms will have only one root. So, the quadratic will have its {discriminant(b^2-4ac)}=0. This will give you the same expression between m and c as you had in the previous condition.Now, solve for the parabola and the line, and using the same logic as above, obtain another relation between m and c.This gives you two relations between the two variables, and thus you will be able to solve them.

The logic is simple. To find the equation of the normal to the given curve at the given point, just find the slope of tangent at the given point. The slope of the normal is just negative of the multiplicative inverse of calculated tangent's slope.The general equation of a line requires, slope and a point that the line passes through; in order to predict the relationship between the x-cor and y-cor, of all the points lying on the line.You have calculated the slope and the point description has been provided in the question, so the equation of the normal is  now known.The coordinates of the desired points can be  calculated easily: since we have two variables to solve for and exactly two equations to solve from, one is the equation of the normal that we have just calculated and other is the equation of the parabola.The point satisfies these two equations since it lies on both the line and the parabola.The answer is, X-cor = -4a/p -2ap                              Y-cor = (x-cor)^2/4a

equation of chord pass the focal point of the parabola y^2=4.ax is y= m(x-a)x coordinates of the intersection of this line to the parabola are the solutions of equation :m^2(x-a)^2= 4a.xor m^2.x^2 -2a(m^2+2)x+ m^2.a^2=0product x1.x2= m^2.a^2/m^2= a^2