The locus of the point -tan-tan- is

The correct option is A xy=1 x=θ+tanθ And #160; y=θ tanθ .Now we know that #160; ^2θ tan ^2θ=1 Or #160; (θ tanθ)(θ+tanθ)=1 Or #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Parametric Representation: Ellipse

The locus of ...

Question

The locus of the point
$(sec θ + tan θ, sec θ - tan θ)$ is

x y = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} = 4 a x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y^{2} = 4 a x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x y = c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x y = 1$
$x = sec θ + tan θ$
And
$y = sec θ - tan θ$ .
Now we know that
${sec}^{2} θ - {tan}^{2} θ = 1$
Or
$(sec θ - tan θ) (sec θ + tan θ) = 1$
Or
$(y) (x) = 1$
Or
$x y = 1$

Suggest Corrections

Similar questions

Q. If chord

P Q

subtends an angle

θ

at the vertex of

y^{2} = 4 a x

, then

tan θ =

x = sec θ - tan θ, y = c o s e c θ + cot θ

Prove:

x y + 1 = y - x

Q. Match the following columns with the values obtained for the solution.

$I .$ $x cos θ + y sin θ = a$ , $x sin θ - y cos θ = b$	$a)$ $(x^{2} - y^{2})^{2} = 16 x y$
$I I .$ $x = sec θ + tan θ$ , $y = sec θ - tan θ$	$b)$ $x y = 1$
$I I I .$ $x sec θ + y tan θ = a$ , $x tan θ + y sec θ = b$	$c)$ $x^{2} - y^{2} = a^{2} - b^{2}$
$I V .$ $x = cot θ + cos θ$ , $y = cot θ - cos θ$	$d)$ $x^{2} + y^{2} = a^{2} + b^{2}$

The locus of the point of intersection of the lines $a x s e c θ + b y t a n θ = a a n d a x t a n θ + b y s e c θ = b$ , where $θ$ is the parameter, is

Q. If the chord of contact of the tangents from a point

P

to the parabola

y^{2} = 4 a x

touches the parabola

x^{2} = 4 b y

, then the locus of

P

Join BYJU'S Learning Program

The locus of the point (secθ+tanθ,secθ−tanθ) is

The correct option is A xy=1x=secθ+tanθAnd y=secθ−tanθ.Now we know that sec2θ−tan2θ=1Or (secθ−tanθ)(secθ+tanθ)=1Or (y)(x)=1Or xy=1

The locus of the point
$(sec θ + tan θ, sec θ - tan θ)$ is

The correct option is A $x y = 1$
$x = sec θ + tan θ$
And
$y = sec θ - tan θ$ .
Now we know that
${sec}^{2} θ - {tan}^{2} θ = 1$
Or
$(sec θ - tan θ) (sec θ + tan θ) = 1$
Or
$(y) (x) = 1$
Or
$x y = 1$