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JEE Mains Maths · Conic Sections

Tangents and Normals to a Parabola

The tangent to a parabola in point, parametric and slope form, the two tangents from an outside point and where tangents meet, and the normal: where it meets the parabola again and how it gives the shortest distance.

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Why this matters

Twenty-five PYQs. The slope form y = mx + a/m answers most tangent questions in one line, and the parametric normal answers the rest. Three ideas cover the page.

Concept 1 of 3: The tangent: point, parametric and slope forms

Substitute a line into y2=4axy^2=4ax and you get a quadratic. The line is a tangent when that quadratic has a double root. For y=mx+cy=mx+c this happens exactly when c=amc=\frac{a}{m}, which is the slope form. The tangent at a known point comes from the T=0T=0 rule; at the parametric point it is ty=x+at2ty=x+at^2.

Definition

  • At (x1,y1)(x_1,y_1): yy1=2a(x+x1)yy_1=2a(x+x_1).
  • At tt: ty=x+at2ty=x+at^2.
  • Slope form: y=mx+amy=mx+\frac{a}{m}, touching at (am2,2am)\left(\frac{a}{m^2},\frac{2a}{m}\right).
  • For x2=4ayx^2=4ay: y=mx−am2y=mx-am^2, touching at (2am, am2)(2am,\,am^2).
  • For y=px2+qx+ry=px^2+qx+r: substitute the line and set the discriminant to zero, or use the derivative.

Slope form for y² = 4ax

y=mx+am,touching at (am2,2am)y=mx+\frac{a}{m},\qquad \text{touching at }\left(\frac{a}{m^2},\frac{2a}{m}\right)

Worked example

Find the tangent to y2=8xy^2=8x with slope 22, and its point of contact.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 3 · Q69Moderate

Example 1 · Conic Sections · Tangents and Normals to a Parabola

A tangent is drawn to the parabola y2=6xy^{2}= 6x which is perpendicular to the line 2x+y=12x+y= 1. Which of the following points does NOT lie on it?

The contact point is (am2,2am)\left(\frac{a}{m^2},\frac{2a}{m}\right), not (a,2a)(a,2a)

(a,2a)(a,2a) is the end of the latus rectum, where the slope is 11. For any other slope use (am2,2am)\left(\frac{a}{m^2},\frac{2a}{m}\right).
Drill 8 more on the tangent: point, parametric and slope forms

Concept 2 of 3: Two tangents: from a point, where they meet, and the directrix

A tangent y=mx+amy=mx+\frac{a}{m} passes through (h,k)(h,k) when hm2−km+a=0hm^2-km+a=0. That quadratic's two roots are the slopes of the two tangents from the point, so their sum and product come free. If the tangents are perpendicular, the product ah\frac{a}{h} is −1-1, so h=−ah=-a: the point is on the directrix. And the tangents at t1t_1 and t2t_2 meet at a point with a clean formula.

Definition

  • Slopes from (h,k)(h,k): roots of hm2−km+a=0hm^2-km+a=0, so m1+m2=khm_1+m_2=\frac{k}{h}, m1m2=ahm_1m_2=\frac{a}{h}.
  • Perpendicular tangents meet on the directrix x=−ax=-a.
  • Tangents at t1,t2t_1,t_2 meet at (at1t2, a(t1+t2))(at_1t_2,\ a(t_1+t_2)).
  • Chord of contact of (h,k)(h,k): ky=2a(x+h)ky=2a(x+h).
  • The tangent at tt meets the axis at (−at2,0)(-at^2,0) and the directrix at (−a, at−at)\left(-a,\ at-\frac{a}{t}\right).

Where the tangents at t₁ and t₂ meet

(at1t2, a(t1+t2))(at_1t_2,\ a(t_1+t_2))

Worked example

Show that the tangents from (−2,1)(-2,1) to y2=8xy^2=8x are perpendicular.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 21 · Q66Moderate

Example 2 · Conic Sections · Tangents and Normals to a Parabola

If two tangents drawn from a point PP to the parabola y2=16(x−3)y^{2}= 16(x- 3) are at right angles, then the locus of point PP is :

Shift before using x=−ax=-a

For y2=16(x−3)y^2=16(x-3) the vertex is at (3,0)(3,0), so the directrix is x=3−4=−1x=3-4=-1, not x=−4x=-4.
Drill 8 more on two tangents: from a point, where they meet, and the directrix

Concept 3 of 3: The normal, and the shortest distance to a parabola

The normal at tt is perpendicular to the tangent ty=x+at2ty=x+at^2, so its slope is −t-t. Written with its slope mm, a normal is y=mx−2am−am3y=mx-2am-am^3, a cubic in mm: up to three normals pass through a point. The shortest distance from an outside point to the parabola is measured along a normal, so to find it, find the normal through that point.

Definition

  • At tt: y=−tx+2at+at3y=-tx+2at+at^3.
  • Slope form: y=mx−2am−am3y=mx-2am-am^3, with foot (am2,−2am)(am^2,-2am).
  • Meets the parabola again at t2=−t−2tt_2=-t-\frac{2}{t}.
  • Three normals from (h,0)(h,0) need h>2ah>2a.
  • Shortest distance from a point: along the normal through it; compare with the vertex if needed.

Normal in slope form, y² = 4ax

y=mx−2am−am3,foot (am2, −2am)y=mx-2am-am^3,\qquad \text{foot }(am^2,\,-2am)

Worked example

The normal to y2=4xy^2=4x at (1,2)(1,2) meets the parabola again at QQ. Find QQ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 24 · Q73Moderate

Example 3 · Conic Sections · Tangents and Normals to a Parabola

Consider the parabola with vertex (12,34)\left( \frac{1}{2},\frac{3}{4} \right) and the directrix y=12y=\frac{1}{2}. Let PP be the point where the parabola meets the line x=−12x = -\frac{1}{2}. If the normal to the parabola at PP intersects the parabola again at the point QQ, then (PQ)2(PQ)^{2} is equal to:

The slope-form foot is (am2,−2am)(am^2,-2am)

For the normal y=mx−2am−am3y=mx-2am-am^3 the foot has a MINUS 2am2am. The tangent's contact point (am2,2am)\left(\frac{a}{m^2},\frac{2a}{m}\right) is a different point with a different mm.
Drill 6 more on the normal, and the shortest distance to a parabola

Summary — formulas & gotchas at a glance

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Formulas (3)

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