Question

If P n Q are two points whose coordinates are (at²,2at) n (a/t²,-2a/t) resp n S is the pt (a,0), then show that 1/SP+1/SQ is independent of t.

Answer 1

SQ = $\sqrt{(a- \frac{a}{t^2})^2 + (0+ \frac{2a}{t})^2}$

= $\sqrt{(\frac{t^2a-a}{t^2})^2 + (\frac{2a}{t})^2}$

= $\sqrt{ \frac{t^4a^2+a^2-2a^2t^2}{t^4} + \frac{4a^2}{t^2} }$

= $\frac{t^4a^2 + a^2 - 2a^2t^2 + 4a^2t^2}{t^4}$

= $\sqrt{ \frac{t^4a^2 + 2a^2t^2 + a^2 }{t^4} }$

= $\sqrt{ \frac{(t^2a+a)^2}{t^4} }$

= $\frac{t^2a+a}{t^2}$

$\frac{1}{SQ} = \frac{t^2 }{t^2a+a}$ ..........(1)

Now, 
SP = $\sqrt{(a-at^2)^2 + (0-2at)^2 }$

= $\sqrt{a^2 - 2a^2t^2 + a^2t^4 + 4a^2t^2}$

= $\sqrt{a^2 + 2a^2t^2 + a^2t^4}$

= $\sqrt{(at^2+a)^2}$

= at² + a

$\frac{1}{SP} = \frac{1}{at^2 + a}$

$\frac{1}{SQ} + \frac{1}{SP} = \frac{t^2 }{t^2a+a}+\frac{1}{at^2 + a}$

= $\frac{t^2+1}{a(t^2+1)}$

= $\frac{1}{SQ} + \frac{1}{SP} = \frac{1 }{a}$

Answer 2

Step-by-step explanation:

We have,

SP=

(at

2

−a)

2

+(2at−0)

2

=a

(t

2

−1)

2

+4t

2

=a(t

2

+1)

and, SQ=

(

t

2

a

−a)

2

+(

t

2a

−0)

2

⇒SQ=

t

4

a

2

(1−t

2

)

2

+

t

2

4a

2

⇒SQ=

t

2

a

(1−t

2

)

2

+4t

2

=

t

2

a

(1+t

2

)

2

=

t

2

a

(1+t

2

)

∴

SP

1

+

SQ

1

=

a(t

2

+1)

1

+

a(t

2

+1)

t

2

⇒

SP

1

+

SQ

1

=

a(t

2

+1)

1+t

2

=

a

1

,whichisindependentoft