True False[1 Marks ]

Question 11 Mark

The line $x + 3y = 0$ is a diameter of the circle $x^2 + y^2 + 6x + 2y = 0.$

Answer

False.
Given equation of the circle is,
$x^2 + y^2 + 6x + 2y = 0$
Centre is $(-3, -1)$
If $x + 3y = 0$ is the equation of diameter, then the centre $(-3, -1)$ will lie on $x + 3y = 0$
$-3 + 3(-1) = 0$
$\Rightarrow -6 \neq 0$
So$, x + 3y = 0$ is not the diameter of the circle.
Hence, he given statement is False.

View full question & answer→

Question 21 Mark

The shortest distance from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ 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.
Given circle is $x^2 + y^2 - 14x - 10y - 151 = 0$
$\therefore$Centre$\ = \text{C}(7,5)$
And 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

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

The line $lx + my + n = 0$ will touch the parabola $y^2 = 4ax$ if ln $= am^2$

Answer

True.
Give line $lx + my + n = 0$ and parabola $y^2 = 4ax$
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.
$\therefore$ 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

If the line $lx + my = 1$ is a tangent to the circle $x^2 + y^2 = a^2$, then the point $\text{(l, m)}$ 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.
Given equation of circle is $x^2 + y^2 = a^2$
and the tangent is $lx + my = 1$
Here centre is $(0, 0)$ and radius $= a$
If $\text{(l, m)}$ 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 $\text{(l, m)}$ lies on the circle.
Hence, the given statement is True.

View full question & answer→

Question 61 Mark

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

The point $(1, 2)$ lies inside the circle $x^2 + y^2 - 2x + 6y + 1 = 0.$

Answer

False.
Given equation of circle is $x^2 + y^2 - 2x + 6y + 1 = 0$
Here $2\ g = -2 \Rightarrow g = -1$
$2f = 6 \Rightarrow f = 3$
$\therefore$ 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

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 ]

Test yourself on this topic