True False[1 Marks ]

Question 11 Mark

State Whether the statements are True or False. Justify.
The line x + 3y = 0 is a diameter of the circle x² + y² + 6x + 2y = 0.

Answer

False.

Solution:

Given equation of the circle is,

x² + y² + 6x + 2y = 0

Centre is (-3, -1)

If x + 3y = 0 is the equation of diameter, then the centre (-3, -1) will lie on x + 3y = 0

-3 + 3(-1) = 0

⇒ -6 ≠ 0

So, x + 3y = 0 is not the diameter of the circle.

Hence, he given statement is False.

View full question & answer→

Question 21 Mark

State Whether the statements are True or False. Justify.
The shortest distance from the point (2, -7) to the circle x² + y² - 14x - 10y - 151 = 0 is equal to 5.
[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point]

Answer

False.

Solution:

Given circle is x² + y² - 14x - 10y - 151 = 0

$\therefore\text{ Centre}\equiv\text{C}(7,5)$

And $\text{Radius}=\sqrt{49+25+151}=\sqrt{225}=15$

Now distance between the point P(2, -7) and centre,

$=\sqrt{(2-7)^2+(-7-5)^2}=\sqrt{25+144}=\sqrt{169}=13$

$\therefore$ Shortest distance of point P from the circle = |13 - 15| = 2

View full question & answer→

Question 31 Mark

State Whether the statements are True or False. Justify.
The locus of the point of intersection of lines $\sqrt{3}\text{x}-\text{y}-4\sqrt{3}\text{k}=0$ and $\sqrt{3}\text{kx}+\text{ky}-4\sqrt{3}=0$ for different value of k is a hyperbola whose eccentricity is 2.
[Hint: Eliminate k between the given equations]

Answer

True.
Solution:
Given equation of lines are,
$\sqrt{3}\text{x}-\text{y}-4\sqrt{3}\text{k}=0\ ...(\text{i})$
and $\sqrt{3}\text{kx}+\text{ky}-4\sqrt{3}=0\ ....(\text{ii})$
From Eq. (i), $\text{k}=\frac{\sqrt{3}\text{x}-\text{y}}{4\sqrt{3}}$
From Eq. (ii), $\text{k}=\frac{4\sqrt{3}}{\sqrt{3}\text{x}+\text{y}}$
Equating the values of k, we get
$\frac{\sqrt{3}\text{x}-\text{y}}{4\sqrt{3}}=\frac{4\sqrt{3}}{\sqrt{3}\text{x}+\text{y}}$
$\Rightarrow3\text{x}^2-\text{y}^2=48$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{48}=1,$ which is equation of hyperbola
$\therefore\ \text{a}^2=16$ and $\text{b}^2=48$
$\Rightarrow\text{e}^2=1+\frac{48}{16}=1+3=4$
$\Rightarrow\text{e}=2$

View full question & answer→

Question 41 Mark

State Whether the statements are True or False. Justify.
The line lx + my + n = 0 will touch the parabola y² = 4ax if ln = am²

Answer

True.

Solution:

Give line lx + my + n = 0 and parabola y² = 4ax

Solving line and parabola for their point of intersection, we get

$\frac{1}{4\text{a}}\text{y}^2+\text{my}+\text{n}=0$

Since line touches the parabola, above equation must have equal roots.

$\therefore\ \text{Discriminant},\text{ D}=0$

$\therefore\text{ m}^2-4\Big(\frac{1}{4\text{a}}\Big)\text{n}=0$

$\Rightarrow\text{ am}^2=\text{nl}$

View full question & answer→

Question 51 Mark

State Whether the statements are True or False. Justify.
If the line lx + my = 1 is a tangent to the circle x² + y² = a², then the point (l, m) lies on a circle.
[Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle]

Answer

True.

Solution:

Given equation of circle is x² + y² = a²

and the tangent is lx + my = 1

Here centre is (0, 0) and radius = a

If (l, m) lie on the circle

$\therefore\ \sqrt{(1-0)^2+(\text{m}-0)^2}=\text{a}$

$\Rightarrow\sqrt{\text{l}^2+\text{m}^2}=\text{a}$

$\Rightarrow\text{l}^2+\text{m}^2=\text{a}^2$ (which is a circle)

So, the point (l, m) lies on the circle.

Hence, the given statement is True.

View full question & answer→

Question 61 Mark

State Whether the statements are True or False. Justify.
The line 2x + 3y = 12 touches the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2$ at the point (3, 2).

Answer

True.

Solution:

If line 2x + 3y = 12 touches the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2,$ at the point (3, 2) satisfies both line and ellipse.

$\therefore$ For line 2x + 3y = 12

2(3) + 3(2) = 12

6 + 6 = 12

12 = 12 True.

For ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=2$

$\frac{(3)^2}{9}+\frac{(2)^2}{4}=2$

$\frac{9}{9}+\frac{4}{4}=2$

$1+1=2$

$2=2\text{ True}$

Hence, the given statement is true.

View full question & answer→

Question 71 Mark

State Whether the statements are True or False. Justify.
The point (1, 2) lies inside the circle x² + y² - 2x + 6y + 1 = 0.

Answer

False.

Solution:

Given equation of circle is x² + y² - 2x + 6y + 1 = 0

Here 2g = -2 ⇒ g = -1

2f = 6 ⇒ f = 3

$\therefore$ Centre = (-g, -f) = (1, -3)

and $\text{r}=\sqrt{\text{g}^2+\text{f}^2-\text{c}}=\sqrt{1+9-1}=3$

$\therefore$ Distance between the point lies outside the circle.

Hence, the given statement is False.

View full question & answer→

Question 81 Mark

State Whether the statements are True or False. Justify.
If P is a point on the ellipse $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{25}=1$ whose foci are S and S', then PS + PS' = 8.

Answer

False.

Solution:

We have equation of the qllipse is $\frac{\text{x}^2}{16}+\frac{\text{y}^2}{25}=1$

From the definition of the ellipse, we know that sum of the distance of any point P on the ellipse from the foci is equal to the length of the major axis.

Here major axis = 2b = 2 × 5 = 10

S and S' are foci, then SP + S'P = 10

View full question & answer→

MATHS True False[1 Marks ]

Take a timed test