#### To determine

**To show:**

PQ1PP1+PQ2PP2=1

#### Answer

PQ1PP1+PQ2PP2=1

#### Explanation

**1) Concept:**

T1 and T2 are drawn at two points P1 and P2 on the parabola y=x2. Slope of tangent is the derivative of the given curve. Using this to find equation of both tangents.

By equating both tangents find point of intersection. By using distance formula find values of PQ1, PQ2, PP1 and |PP2|.

**2) Formula:**

(i) Slope of tangent =

mtanΑ=dydx

(ii) Equation of tangent through a, b is

y-b=mtanΑx-a

(iii) Distance between two points x1, y1 and (x2, y2) is given by

d=x2-x12+y2-y12

**3) Given:**

Tangent lines T1 and T2 drawn at points P1 and P2 on parabola y=x2, intersect at a point P. Another tangent line is drawn at a point between P1 and P2, which intersects T1 at Q1 and T2 at Q2.

**4) Calculations:**

Slope of tangent to the curve y=x2 is dydx=2x,

Therefore general equations of tangent to the curve at point Say P1a, a2 and P2b,b2, are

y-a2=2ax-a that is y=2ax-a2 and y=2bx-b2,

At point of intersection P, 2ax-a2=2bx-b2

Therefore, 2a-bx= a2-b2

Hence, x=a+b2,

Using this in any one equation, y =2aa+b2-a2=ab.

Therefore, point of intersection is,Pa+b2, ab.

Let Q be point on the curve between P1 and P2 with co-ordinates, say c, fc. Let T be a tangent to the curve at Q.

Let tangent T and T`1 intersects at point Q1 then from above it follows that the coordinates of Q1 is given by

a+c2, ac similarly

tangent T and T2 intersects at point Q2 with co-ordinates

b+c2, bc

Therefore, by using distance formula,

PQ1=a+b2-a+c22+ab-ac2

=b-c22+a2b-c2

=b-c14+a2

PQ2=a+b2-b+c22+ab-bc2

=a-c22+b2a-c2

=a-c14+b2

PP1=a+b2-a2+ab-a22

=b-a22+a2b-a2

=b-a14+a2

PP2=a+b2-b2+ab-b22

=a-b22+b2a-b2

=a-b14+b2

Therefore,

PQ1PP1=b-c14+a2b-a14+a2=(b-c)(b-a)

PQ2PP2=a-c14+b2a-b14+b2=a-ca-b

PQ1PP1+PQ2PP2=(c-b)(a-b)+a-ca-b=a-ba-b=1

**Conclusion:**

PQ1PP1+PQ2PP2=1