First slide

Parabola

Question

$T h e f o c u s o f t h e p a r a b o l a i s (1, 1) a n d t h e \tan g e n t a t t h e v e r t e x h a s t h e e q u a t i o n x + y = 1 w h i c h o f t h e f o l l o w i n g i s c o r r e c t$

Difficult

A

Equation of parabola is ${(x - y)}^{2} = 2 (x + y - 1)$
B

Equation of parabola is ${(x - y)}^{2} = 4 (x + y - 1)$
C

The coordinates of the vertex $(\frac{1}{2}, \frac{1}{2})$
D

The length of the latursectum is $2 \sqrt{2}$

Solution

$T h e d i s \tan c e f r o m f o c u s t o t h e \tan g e n t a t v e r t e x i s a = \sqrt{2}$
$D i r e c t r i x i s a l i n e p a r a l l e l t o t h e \tan g e n t a t v e r t e x a n d a t a d i s \tan c e o f 2 \sqrt{2} u n i t s f r o m f o c u s$
$D i r e c t r i x i s p a s \sin g t h r o u g h t h e i m a g e o f f o c u s i n \tan g e n t a t v e r t e x$
$H e n c e t h e e q u a t i o n o f t h e d i r e c t r i x i s x + y = 0$
$B y u \sin g t h e d e f i n i t i o n o f t h e p a r a b o l a, t h e l o c u s i s$
${(\frac{x + y}{\sqrt{2}})}^{2} = {(x - 1)}^{2} + {(y - 1)}^{2}$
$x^{2} + y^{2} + 2 x y = 2 x^{2} + 2 y^{2} - 4 x - 4 y + 4 = 0$
${(x - y)}^{2} = 4 (x + y - 1)$
$T h i s i s t h e e q u a t i o n o f t h e p a r a b o l a$
$V e r t e x i s t h e f o o t o f t h e p e r p e n d i c u l a r o f f o c u s o n t h e \tan g e n t a t v e r t e x$
$h e r e v e r t e x i s (\frac{1}{2}, \frac{1}{2})$
$a = \frac{1}{\sqrt{2}}, h e n c e t h e l e n g t h o f t h e l a t u s r e c t u m i s 2 \sqrt{2}$

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