# LOCUS & PARABOLAS: See questions below?

P(16,16) and Q (-4,1) are points on the parabola x^2=16y

(i) Show that PQ is a focal chord

(ii) Find the equations of the tangents at P and Q

(iii) Show that these 2 tangents are perpendicular

(iv) Show that the 2 tangents intersect on the directrix

Full working put, please. And thanks for helping out!

### 2 Answers

- JupiterSkyLv 62 years agoFavorite Answer
(i)

First find the focal point of the parabola y = x²/16

Focus: (h,k+1/4a)

In vertex form y = a(x−h)²+k, the parabola in question is expressed as y = ¹/₁₆(x−0)²+0,

so the vertex (h,k) is at (0,0) and a=¹/₁₆

Focus: (0,0+1/4a) = (0,4)

Let F be the focus. If PQ is a focal chord, then PQ has the same slope as PF (or QF)

Slope PQ: ∆y/∆x = ¾

Slope PF: ∆y/∆x = ¾

Slope QF: ∆y/∆x = ¾

-----------> PQ and PF (and QF) have the same slope which shows that PQ is a focal chord

(ii)

dy/dx = x/8

The derivative at the point is the slope at that point:

Slope at P: 16/8 = 2

Slope at Q: −½

(y−y₀) = m(x−x₀)

y = mx−mx₀+y₀

----------->

Tangent line at P: y = 2x−2*16+16 = 2x−16

Tangent line at Q: y = −½x−(−½)*(−4)+1 = −½x−1

(iii)

Two perpendicular lines satisfy the condition: m₁m₂ = −1

Slope P = 2

Slope Q = −½

2 * −½ = −1

-----------> Tangent lines of P and Q are perpendicular since the product of their slopes equals −1

(iv)

The vertex is the midpoint of the line joining the focus and the intersection of the directrix with

the axis of symmetry.

Vertex: (0,0)

Focus: (0,4)

Axis of symmetry: x=0

Intersection of directrix and axis of symmetry is then (0,−4), if (0,0) is the midpoint

=> Directrix: y = −4

The tangent lines of P and Q intersect on the directrix IFF they both have the same

x-coordinate at y = −4

Tangent at P: −4 = 2x−16 => x = 6

Tangent at Q: −4 = −½x−1 => x = 6

-----------> Tangent lines of P and Q intersect on the directrix at the point (6,−4)

How about the other questions?