Conic Sections (Parabola, Ellipse, Hyperbola): some basic theorems/questions: IITJEE Mains Maths

Problem 1:

(A) Proposition 1: Show  that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Proof 1:

Let t_{1} and t_{2} be the extremities of a focal chord of the parabola y^{2}=4ax. Then, it can be shown that t_{1}t_{2}=-1. (Try this part on your own and let me know; if you can’t produce the proof, I will send it to you.) The equations of the tangents at t_{1} and t_{2} are t_{1}y=x+at_{1}^{2} and t_{2}y=x+at_{2}^{2}. The product of the slopes is \frac{1}{t_{1}}-\frac{1}{t_{2}}=-1

Therefore, the tangents are at right angles. Also, the point of intersection of  these tangents is x=at_{1}t_{2}, y=a(t_{1}+t_{2}), that is, x=-a, y=a(t_{1}+t_{2}), which clearly lies on the directrix x=-a.

(B) Proposition 2: The tangent at any point of a parabola bisects the angle between the focal distance of the point and the perpendicular on the directrix from that point. Homework !

(C) Proposition 3: The portion of a tangent to a parabola cut off between the directrix and the curve, subtends a right angle at the focus. Homework!

Problem 2:


Show that in general, three normals can be drawn to a given parabola from a given point, one of which is always real. Also, show that the sum of the ordinates of the feet of these co-normal points is zero.


Let y^{2}=4ax be the given parabola. The equation of a normal to this parabola at (am^{2},-2am) is y=mx-2am-am^{2}. If it passes through a given point (\alpha, \beta) then

\beta = m\alpha -2am-am^{3}, or am^{3}+(2a-\alpha)m + \beta = 0…call this Equation (I)

which, being cubic in m, gives three values of m, say m_{1}. m_{2}, and m_{3}, and hence, three points on the parabola, the normals at which pass through (\alpha, \beta). Since the complex roots of the equation with real coefficients occur in pairs and the degree of the above equation is odd, at least one of the roots is real so there is at least one real normal to the parabola passing through the given point (\alpha, \beta).

From (I), we have

m_{1}+m_{2}+m_{3}=0 \Longrightarrow -2am_{1}-2am_{2}-2am_{3}=0 \Longrightarrow y_{1}+y_{2}+y_{3}=0, where y_{i}=-2am_{i} for i=1, 2, 3. Hence, the result.

Problem 3:

If the tangents and normals at the extremities of a focal chord of the parabola y^{2}=4ax intersect at (x_{1},y_{1}) and (x_{2},y_{2}) respectively, then show that y_{1}=y_{2}.

Answer 3:

Let P(at_{1}^{2}, 2at_{1}) and Q(at_{2}^{2},2at_{2}) be the extremities of a focal chord of the parabola y^{2}=4ax, then

t_{1}t_{2}=-1….call this Relation (I).

Next, equations of the tangents at P and Q are respectively, t_{1}y=x+at_{1}^{2} and t_{2}y=x+at_{2}^{2}

Solving these equations, we get

(t_{1}-t_{2})y=a(t_{1}^{2}-t_{2}^{2}) \Longrightarrow y=a(t_{1}+t_{2}) and


So that x_{1}=-a and y_{1}=a(t_{1}+t_{2})….call this Relation (II).

Now, equations of the normals at P and Q are respectively

y = -t_{1}x+2at_{1}+at_{1}^{3} and y=-t_{2}x+2at_{2}+at_{2}^{3}

Solving these equations we get


\Longrightarrow x =2a +a(t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}) = a(t_{1}^{2}+t_{2}^{2}+1) (using relation I)

and y = -t_{1}a(t_{1}^{2}+t_{2}^{2}+1) + 2at_{1}+at_{1}^{3} = -at_{1}t_{2}^{2}+at_{1}=a(t_{1}+t_{2}) (again by using relationship I),

So that x_{2}=a(t_{1}^{2}+t_{2}^{2}+1) and y_{2}=a(t_{1}+t_{2}) so that we get y_{1}=y_{2}.


Results or relations I and II should be memorized as they are frequently used in the theory and applications.

More later,

Nalin Pithwa.


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